CLASS 11 PHYSICS | MOTION IN A PLANE

Motion in a Plane and Relative Velocity

A full-width Physics resource for CBSE, NEET, JEE Main, JEE Advanced, IB, ICSE, IGCSE and A-Level students, with original questions on vectors, river-boat motion, rain-man problems and airplane wind correction.

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v = v_x i + v_y j independent components

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Two-Dimensional Motion, Vectors and Components

Motion in a plane needs two coordinates. Instead of treating the path as one complicated curve, Physics splits each vector into perpendicular components. The x and y components are solved independently and then recombined by vector addition.

Position Vector

The position vector joins the origin to the point occupied by the particle. For point P(x, y), it stores both distance and direction from the origin.

r = x i + y j |r| = √(x2 + y2) tan θ = y / x

Velocity Vector

Velocity is the rate of change of position. Its components tell how fast the particle moves along each axis.

v = vx i + vy j v = √(vx2 + vy2) tan θ = vy / vx

Acceleration Vector

Acceleration is the rate of change of velocity. A body can accelerate even at constant speed if its direction changes.

a = ax i + ay j a = √(ax2 + ay2)

Exam Tip: Decide signs first. Most mistakes in vector problems come from subtracting components with inconsistent directions.

P(x,y) x component y component r

Component Method

Resolve every vector into x and y parts, operate on like components, then recombine the final components.

Vector Subtraction

For relative velocity, subtract the observer's velocity from the object's velocity. Direction matters as much as magnitude.

Observer Dependence

The same motion can look different from ground, train, boat, airplane or rain-frame observers.

Relative Velocity in Two Dimensions

Relative velocity answers one question: how does one object appear to move for an observer moving with another object? If A is observed from B, subtract B's velocity from A's velocity.

General Formula

vAB = vA - vB

Use this in component form for all directions.

Same Line

Same direction gives speed difference. Opposite direction gives speed sum.

Oblique Motion

Draw a velocity triangle or use components. Avoid adding speeds directly unless directions are the same.

Exam Tip: The word "with respect to" names the observer. In vAB, B is the observer, so vB is subtracted.

River-Boat Problems: Shortest Time, Shortest Path and No Drift

River current acts along the river. Boat velocity relative to water is controlled by the boatman. The velocity relative to ground is the vector sum of these two velocities.

Shortest Time

Use maximum across-river component. The boat is aimed perpendicular to the bank.

tmin = width / udrift = vrivert

Shortest Path

The resultant ground velocity must be perpendicular to the bank, so downstream drift is zero.

u sin θ = vrivert = width / √(u2 - vriver2)

Exam Tip: Shortest time and shortest path are usually different. Shortest time allows drift; shortest path cancels drift.

boat across river current ground velocity

Rain-Man and Airplane-Wind Problems

Rain, umbrella and aircraft questions are the same vector idea in different language. Apparent rain is rain velocity relative to the observer. Aircraft ground velocity is aircraft velocity relative to air plus wind velocity.

Rain-Man Problems

For an observer moving with velocity vman, apparent rain velocity is:

vrain,man = vrain - vman

The umbrella is held along the apparent rain direction.

Airplane-Wind Problems

Wind is added to aircraft velocity relative to air to get ground velocity:

vground = vaircraft,air + vwind

In a crosswind, the aircraft must point into the wind to keep the desired ground track.

observer apparent rain
ground track heading wind

Exam Tip: Do not memorize separate tricks for rain, boats and airplanes. Write the velocities in the chosen frame and add or subtract vectors.

NEET Motion in a Plane Questions

These 25 NEET-level MCQs are original, non-duplicate and focused on fast conceptual accuracy with clean calculations.

NEET 1: Resultant velocity from rectangular components

Question: A particle has velocity components 6 m s-1 east and 8 m s-1 north. What is its speed and direction?

  1. 10 m s-1, tan θ = 4/3 north of east
  2. 14 m s-1, tan θ = 3/4 north of east
  3. 2 m s-1, along north
  4. 48 m s-1, along east

Correct Answer: A. 10 m s-1, tan θ = 4/3 north of east

Short Explanation: Perpendicular components combine as v = √(62 + 82) = 10 m s-1. The angle from east satisfies tan θ = vy/vx = 8/6.

NEET 2: Finding components from speed and angle

Question: A velocity of 20 m s-1 makes 30° with the positive x-axis. Which pair gives its x and y components?

  1. 10√3, 10
  2. 10, 10√3
  3. 20, 20
  4. 5√3, 15

Correct Answer: A. vx = 10√3 m s-1, vy = 10 m s-1

Short Explanation: Use vx = v cos θ and vy = v sin θ. With θ = 30°, vx = 20(√3/2) and vy = 20(1/2).

NEET 3: Relative velocity of perpendicular vehicles

Question: Car A moves east at 12 m s-1 and car B moves north at 5 m s-1. The speed of A relative to B is

  1. 7 m s-1
  2. 13 m s-1
  3. 17 m s-1
  4. 60 m s-1

Correct Answer: B. 13 m s-1

Short Explanation: vAB = vA - vB = 12 i - 5 j. Its magnitude is √(122 + 52) = 13 m s-1.

NEET 4: River crossing in minimum time

Question: A river is 120 m wide. A boat has speed 5 m s-1 in still water and the river current is 3 m s-1. If the boat is aimed straight across, the crossing time is

  1. 15 s
  2. 24 s
  3. 30 s
  4. 40 s

Correct Answer: B. 24 s

Short Explanation: Minimum time is obtained by using the full boat speed across the river. Therefore t = width / boat speed = 120/5 = 24 s. The current only produces drift.

NEET 5: Drift during shortest-time river crossing

Question: In the previous type of crossing, a boat is aimed perpendicular to a 90 m wide river. Boat speed is 6 m s-1 and current speed is 2 m s-1. The downstream drift is

  1. 15 m
  2. 20 m
  3. 30 m
  4. 45 m

Correct Answer: C. 30 m

Short Explanation: The crossing time is 90/6 = 15 s. Drift = current speed x time = 2 x 15 = 30 m downstream.

NEET 6: No-drift river condition

Question: A boat of speed 10 m s-1 must cross a river of current 6 m s-1 without drift. The upstream component of boat velocity must be

  1. 4 m s-1
  2. 6 m s-1
  3. 8 m s-1
  4. 10 m s-1

Correct Answer: B. 6 m s-1

Short Explanation: For no drift, the upstream component of boat velocity relative to water must exactly cancel the river current. Hence vboat,upstream = 6 m s-1.

NEET 7: Impossible straight crossing

Question: A boat can move at 4 m s-1 in still water while the river current is 6 m s-1. Which statement is correct?

  1. The boat can reach exactly opposite by aiming upstream
  2. The boat cannot cancel the current fully
  3. The crossing time is zero
  4. The resultant velocity can be purely upstream

Correct Answer: B. The boat cannot cancel the current fully

Short Explanation: The maximum upstream component available is 4 m s-1, which is less than the 6 m s-1 current. A no-drift path is physically impossible.

NEET 8: Rain seen by a walking observer

Question: Rain falls vertically downward at 10 m s-1. A student walks east at 6 m s-1. The rain appears to make an angle with the vertical such that

  1. tan θ = 3/5
  2. tan θ = 5/3
  3. tan θ = 4/3
  4. tan θ = 0

Correct Answer: A. tan θ = 3/5

Short Explanation: Relative rain velocity has horizontal component 6 m s-1 opposite the walking direction and vertical component 10 m s-1. Thus tan θ = 6/10 = 3/5.

NEET 9: Umbrella held vertical

Question: A man moves east at 4 m s-1. For rain to appear vertical to him, the actual rain must have which horizontal component?

  1. 4 m s-1 east
  2. 4 m s-1 west
  3. 0 m s-1
  4. 8 m s-1 west

Correct Answer: A. 4 m s-1 east

Short Explanation: The apparent horizontal component is vrain,x - vman,x. For it to be zero, vrain,x must equal the man's eastward speed.

NEET 10: Aircraft heading in crosswind

Question: An aircraft has airspeed 200 km h-1. A wind blows east at 40 km h-1. To fly due north over the ground, the aircraft should head

  1. Directly north
  2. West of north with sin θ = 0.2
  3. East of north with sin θ = 0.2
  4. South of east

Correct Answer: B. West of north with sin θ = 0.2

Short Explanation: The aircraft must provide a 40 km h-1 westward component to cancel the eastward wind. Hence 200 sin θ = 40.

NEET 11: Tailwind and ground speed

Question: A plane flies with airspeed 160 km h-1 in a 20 km h-1 tailwind. Its ground speed is

  1. 140 km h-1
  2. 160 km h-1
  3. 180 km h-1
  4. 3200 km h-1

Correct Answer: C. 180 km h-1

Short Explanation: A tailwind is in the same direction as the aircraft's air velocity, so the velocity relative to ground is 160 + 20 = 180 km h-1.

NEET 12: Headwind travel time

Question: A plane has airspeed 180 km h-1 and flies 540 km against a 30 km h-1 headwind. The time taken is

  1. 2 h
  2. 3 h
  3. 3.6 h
  4. 4.5 h

Correct Answer: C. 3.6 h

Short Explanation: Ground speed against the wind is 180 - 30 = 150 km h-1. Time = 540/150 = 3.6 h.

NEET 13: Same-direction relative speed

Question: Two cyclists move along the same straight road in the same direction at 20 m s-1 and 15 m s-1. Their relative speed is

  1. 5 m s-1
  2. 15 m s-1
  3. 20 m s-1
  4. 35 m s-1

Correct Answer: A. 5 m s-1

Short Explanation: For motion in the same direction, relative speed is the difference of speeds: 20 - 15 = 5 m s-1.

NEET 14: Opposite-direction relative speed

Question: Two particles move in opposite directions along the same line at 9 m s-1 and 11 m s-1. Their relative speed is

  1. 2 m s-1
  2. 9 m s-1
  3. 11 m s-1
  4. 20 m s-1

Correct Answer: D. 20 m s-1

Short Explanation: Opposite directions make the separation change at the sum of speeds, so relative speed = 9 + 11 = 20 m s-1.

NEET 15: Quadrant of resultant velocity

Question: A velocity vector is v = -3 i + 4 j m s-1. Its direction lies in which quadrant?

  1. First quadrant
  2. Second quadrant
  3. Third quadrant
  4. Fourth quadrant

Correct Answer: B. Second quadrant

Short Explanation: The x-component is negative and the y-component is positive. Therefore the vector points left and upward, which is the second quadrant.

NEET 16: Resultant boat velocity

Question: A boat is aimed straight across a river with speed 6 m s-1. The current is 8 m s-1. The magnitude of its ground velocity is

  1. 2 m s-1
  2. 7 m s-1
  3. 10 m s-1
  4. 14 m s-1

Correct Answer: C. 10 m s-1

Short Explanation: Across and downstream components are perpendicular. Ground speed = √(62 + 82) = 10 m s-1.

NEET 17: Relative velocity vector components

Question: For vA = 7 i + 2 j and vB = 1 i - 3 j, the velocity of A with respect to B is

  1. 6 i + 5 j
  2. 8 i - 1 j
  3. -6 i - 5 j
  4. 7 i - 3 j

Correct Answer: A. 6 i + 5 j

Short Explanation: Use vAB = vA - vB. Therefore vAB = (7 - 1)i + (2 - (-3))j = 6 i + 5 j.

NEET 18: Observer-dependent velocity

Question: A passenger in a train sees a pole moving backward although the pole is at rest on the ground. The best explanation is

  1. Velocity depends on the observer's frame
  2. The pole really moves backward
  3. Acceleration is always zero
  4. Relative velocity is not a vector

Correct Answer: A. Velocity depends on the observer's frame

Short Explanation: In the train frame, the ground and fixed pole have velocity opposite to the train. This is a direct use of relative velocity.

NEET 19: Acceleration from components

Question: An acceleration vector has components 3 m s-2 and 4 m s-2 along perpendicular axes. Its magnitude is

  1. 1 m s-2
  2. 5 m s-2
  3. 7 m s-2
  4. 12 m s-2

Correct Answer: B. 5 m s-2

Short Explanation: The magnitude of a vector from perpendicular components is √(32 + 42) = 5 m s-2.

NEET 20: Displacement vector magnitude

Question: A displacement is 5 i + 12 j metres. Its magnitude is

  1. 7 m
  2. 13 m
  3. 17 m
  4. 60 m

Correct Answer: B. 13 m

Short Explanation: Magnitude = √(52 + 122) = √169 = 13 m.

NEET 21: Direction of a position vector

Question: The position vector of a point is r = 4 i + 4 j. The angle made with the positive x-axis is

  1. 0°
  2. 30°
  3. 45°
  4. 90°

Correct Answer: C. 45°

Short Explanation: tan θ = y/x = 4/4 = 1. Since both components are positive, θ = 45° in the first quadrant.

NEET 22: Shortest path condition in a river

Question: For a boat to take the shortest path across a river, its ground velocity must be

  1. Along the river
  2. Perpendicular to the banks
  3. Opposite to the boat velocity
  4. Equal to zero

Correct Answer: B. Perpendicular to the banks

Short Explanation: The shortest path between opposite banks is a straight line normal to the river banks. This requires zero downstream drift.

NEET 23: Current and shortest crossing time

Question: While crossing a river, the current affects the shortest crossing time only if it changes

  1. The across-river component of boat velocity
  2. The downstream drift
  3. The river width
  4. The unit vector notation

Correct Answer: A. The across-river component of boat velocity

Short Explanation: Time to cross is width divided by across-river velocity. A uniform current along the river changes drift, not the across component when the boat is aimed perpendicular.

NEET 24: Crosswind and airplane track

Question: A crosswind acts perpendicular to the desired ground track of an aircraft. The pilot should correct mainly by changing

  1. Aircraft mass
  2. Heading direction
  3. Value of g
  4. Passenger speed inside the aircraft

Correct Answer: B. Heading direction

Short Explanation: The aircraft must aim slightly into the wind so that the wind component is cancelled and the ground velocity follows the desired track.

NEET 25: Zero relative velocity

Question: If the relative velocity of A with respect to B is zero, then

  1. A and B have equal velocity vectors
  2. A and B must be at the same point
  3. A has zero speed
  4. B has zero acceleration

Correct Answer: A. A and B have equal velocity vectors

Short Explanation: vAB = vA - vB. If it is zero, the two velocity vectors are equal in magnitude and direction.

JEE Main Motion in a Plane Questions

These 25 JEE Main-level questions include numericals, vector subtraction, river crossing, rain-man and airplane wind correction.

JEE Main 1: Displacement from constant vector velocity

Question: A particle moves with constant velocity v = 3 i + 4 j m s-1 for 5 s. Find its displacement and distance travelled.

Key Formula: s = v t; distance = |v|t for constant straight-line motion.

  1. Displacement = (3 i + 4 j) x 5 = 15 i + 20 j m.
  2. Speed = √(32 + 42) = 5 m s-1.
  3. Distance travelled = 5 x 5 = 25 m.

Answer: Displacement = 15 i + 20 j m; distance = 25 m.

JEE Main 2: Magnitude of relative velocity in 2D

Question: Two particles have velocities vA = 8 i + 6 j and vB = 2 i - 2 j m s-1. Find |vAB|.

Key Formula: vAB = vA - vB.

  1. vAB = (8 - 2)i + (6 - (-2))j = 6 i + 8 j.
  2. Magnitude = √(62 + 82) = 10 m s-1.

Answer: 10 m s-1.

JEE Main 3: River shortest time with drift

Question: A river is 150 m wide. Boat speed in still water is 10 m s-1 and current speed is 6 m s-1. The boat is aimed normal to the bank. Find time and drift.

Key Formula: t = d/u; drift = vcurrentt.

  1. The across component is the full boat speed, 10 m s-1.
  2. t = 150/10 = 15 s.
  3. Drift = 6 x 15 = 90 m downstream.

Answer: 15 s and 90 m downstream.

JEE Main 4: River shortest path with no drift

Question: A boat of speed 10 m s-1 crosses a 96 m wide river with current 6 m s-1. Find the crossing time for reaching the exactly opposite point.

Key Formula: u sin θ = vriver; across speed = √(u2 - vriver2).

  1. The upstream component must be 6 m s-1.
  2. Across component = √(102 - 62) = 8 m s-1.
  3. Time = 96/8 = 12 s.

Answer: 12 s.

JEE Main 5: Umbrella angle in vertical rain

Question: Rain falls vertically at 12 m s-1. A man runs horizontally at 5 m s-1. At what angle with the vertical should he hold the umbrella?

Key Formula: tan θ = vman/vrain.

  1. Relative rain has horizontal component 5 m s-1 opposite to the man.
  2. Its vertical component is 12 m s-1 downward.
  3. Therefore tan θ = 5/12.

Answer: θ = tan-1(5/12), tilted forward in the direction of motion.

JEE Main 6: Airplane wind correction and ground speed

Question: An aircraft has airspeed 250 km h-1. Wind blows east at 50 km h-1. The pilot wants a due-north ground track. Find heading angle and ground speed.

Key Formula: vair sin θ = vwind; vground = vair cos θ.

  1. The aircraft must head west of north to cancel the east wind.
  2. sin θ = 50/250 = 0.2, so θ = sin-1(0.2).
  3. Ground speed = √(2502 - 502) = 100√6 km h-1.

Answer: Head west of north by sin-1(0.2); ground speed = 100√6 km h-1.

JEE Main 7: Closest approach of two moving particles

Question: Particle A starts at the origin and moves north at 10 m s-1. Particle B starts 100 m east of A and moves west at 5 m s-1. Find the time of closest approach.

Key Formula: For relative position r and relative velocity v, tmin = - (r.v)/|v|2.

  1. Position of B relative to A initially is r = 100 i.
  2. Velocity of B relative to A is v = -5 i - 10 j.
  3. tmin = -[100 i . (-5 i - 10 j)]/(52 + 102) = 500/125 = 4 s.

Answer: 4 s.

JEE Main 8: Direction from velocity components

Question: A particle has velocity 5 i + 12 j m s-1. Find its speed and direction with the x-axis.

Key Formula: v = √(vx2 + vy2); tan θ = vy/vx.

  1. Speed = √(52 + 122) = 13 m s-1.
  2. tan θ = 12/5.
  3. Both components are positive, so the angle is measured above +x.

Answer: 13 m s-1, θ = tan-1(12/5).

JEE Main 9: Acceleration from changing components

Question: Velocity changes from 2 i + 3 j to 8 i - 1 j m s-1 in 2 s. Find average acceleration.

Key Formula: aavg = Δv / Δt.

  1. Δv = (8 - 2)i + (-1 - 3)j = 6 i - 4 j.
  2. aavg = (6 i - 4 j)/2 = 3 i - 2 j m s-2.

Answer: 3 i - 2 j m s-2.

JEE Main 10: Condition for one body to appear at rest

Question: For an observer moving with B, particle A appears at rest. What relation must hold between vA and vB?

Key Formula: vAB = vA - vB.

  1. If A appears at rest to B, vAB = 0.
  2. Thus vA - vB = 0.
  3. The velocity vectors must be identical.

Answer: vA = vB.

JEE Main 11: Boat aimed upstream at a given angle

Question: A boat of speed 10 m s-1 is aimed 30° upstream from the normal to the river bank. Current speed is 4 m s-1 and river width is 100 m. Find drift.

Key Formula: Across component = u cos θ; net downstream component = v - u sin θ.

  1. Across component = 10 cos 30° = 5√3 m s-1.
  2. Net downstream component = 4 - 10 sin 30° = -1 m s-1, so drift is upstream.
  3. Time = 100/(5√3) = 20/√3 s. Drift = -20/√3 m.

Answer: 20/√3 m upstream.

JEE Main 12: Rain actually inclined but apparently vertical

Question: Rain falls with speed 20 m s-1 at 30° to the vertical toward the east. What eastward speed should a man have so that rain appears vertical?

Key Formula: For apparent vertical rain, horizontal components of rain and observer are equal.

  1. Rain's horizontal component = 20 sin 30° = 10 m s-1 east.
  2. The man must move east with the same horizontal component.
  3. Then relative horizontal rain velocity becomes zero.

Answer: 10 m s-1 east.

JEE Main 13: Round trip in wind

Question: An aircraft has airspeed 150 km h-1. Wind along the route is 30 km h-1. Find time for a 300 km outward trip with tailwind and 300 km return against wind.

Key Formula: t = d/(u+v) + d/(u-v).

  1. Outward ground speed = 150 + 30 = 180 km h-1.
  2. Return ground speed = 150 - 30 = 120 km h-1.
  3. Total time = 300/180 + 300/120 = 5/3 + 5/2 = 25/6 h.

Answer: 25/6 h, or about 4.17 h.

JEE Main 14: Perpendicular relative velocities

Question: Two particles move with speeds 9 m s-1 and 12 m s-1 along mutually perpendicular directions. Find their relative speed.

Key Formula: vrel = √(u2 + v2) for perpendicular velocities.

  1. Choose one direction as x and the other as y.
  2. Relative velocity components are 9 and 12 in perpendicular directions.
  3. Magnitude = √(92 + 122) = 15 m s-1.

Answer: 15 m s-1.

JEE Main 15: Position vector in the second quadrant

Question: A point has coordinates (-8 m, 6 m). Find the magnitude of its position vector and the angle from +x axis.

Key Formula: |r| = √(x2 + y2); use quadrant after finding tan θ.

  1. Magnitude = √((-8)2 + 62) = 10 m.
  2. The reference angle is tan-1(6/8).
  3. Since x is negative and y is positive, the vector lies in quadrant II.

Answer: 10 m; angle = 180° - tan-1(3/4).

JEE Main 16: Velocity from component equations

Question: A particle has x = 4t and y = 3t + 2, where x and y are in metres and t in seconds. Find its velocity vector.

Key Formula: v = (dx/dt)i + (dy/dt)j.

  1. dx/dt = 4 m s-1.
  2. dy/dt = 3 m s-1.
  3. So v = 4 i + 3 j m s-1.

Answer: 4 i + 3 j m s-1.

JEE Main 17: Minimum drift when current is stronger

Question: A boat has speed 5 m s-1 in still water while river current is 13 m s-1. If the river is 60 m wide, find the least possible downstream drift.

Key Formula: For minimum drift with current stronger than boat, aim as far upstream as possible while still crossing; limiting result gives drift = d√(v2-u2)/u.

  1. Here u = 5 and v = 13, so no-drift crossing is impossible.
  2. At the optimum, the ground velocity is tangent to the reachable direction cone.
  3. Minimum drift = 60 x √(132 - 52)/5 = 60 x 12/5 = 144 m.

Answer: 144 m downstream.

JEE Main 18: Separation on perpendicular roads

Question: Two students start from the same crossing. One walks east at 3 m s-1 and the other north at 4 m s-1. Find their separation after 10 s.

Key Formula: Separation = |vrel|t when initial separation is zero.

  1. Relative speed = √(32 + 42) = 5 m s-1.
  2. After 10 s, separation = 5 x 10 = 50 m.

Answer: 50 m.

JEE Main 19: Relative direction of moving particles

Question: A has velocity 4 i + 2 j and B has velocity -2 i + 10 j m s-1. Find the direction of A relative to B.

Key Formula: vAB = vA - vB; tan θ = component ratio with quadrant.

  1. vAB = (4 - (-2))i + (2 - 10)j = 6 i - 8 j.
  2. The vector lies in the fourth quadrant.
  3. tan of the angle below +x is 8/6 = 4/3.

Answer: tan-1(4/3) below the +x axis.

JEE Main 20: Relative velocity of rain with horizontal wind

Question: Rain has velocity 3 i - 4 j m s-1 relative to ground. A cyclist moves at 8 i m s-1. Find rain velocity relative to cyclist.

Key Formula: vrain,cycle = vrain - vcycle.

  1. vrain,cycle = (3 i - 4 j) - 8 i.
  2. Therefore vrain,cycle = -5 i - 4 j m s-1.
  3. The apparent rain comes from the forward side of the cyclist.

Answer: -5 i - 4 j m s-1.

JEE Main 21: Acceleration from time-dependent velocity

Question: Velocity of a particle is v = (2t + 1)i + (6 - 3t)j m s-1. Find acceleration at t = 2 s.

Key Formula: a = dv/dt.

  1. Differentiate each component separately.
  2. ax = d(2t + 1)/dt = 2.
  3. ay = d(6 - 3t)/dt = -3.

Answer: 2 i - 3 j m s-2.

JEE Main 22: Independence of components

Question: A particle has zero acceleration along x but constant acceleration along y. Which statement is correct?

Key Formula: Two perpendicular components of motion are treated independently.

  1. The x-velocity remains constant because ax = 0.
  2. The y-velocity changes linearly with time because ay is constant.
  3. The total path may still be curved.

Answer: x-motion is uniform while y-motion is uniformly accelerated.

JEE Main 23: No-drift angle and time

Question: A boat of speed 13 m s-1 crosses a river of current 5 m s-1. Width is 240 m. Find time for no drift.

Key Formula: Across component = √(u2 - v2).

  1. The upstream component required is 5 m s-1.
  2. Across component = √(132 - 52) = 12 m s-1.
  3. Time = 240/12 = 20 s.

Answer: 20 s.

JEE Main 24: Velocity triangle for downstream landing

Question: A boat crosses a 100 m wide river and lands 40 m downstream in 20 s. If current is 3 m s-1, find the boat velocity relative to water.

Key Formula: Ground velocity components are displacement/time; boat velocity = ground velocity - river velocity.

  1. Across ground component = 100/20 = 5 m s-1.
  2. Downstream ground component = 40/20 = 2 m s-1.
  3. Boat's downstream component relative to water = 2 - 3 = -1 m s-1.

Answer: 5 j - 1 i m s-1 if downstream is +i and across is +j.

JEE Main 25: Interception by relative motion

Question: A boat starts from origin with speed 10 m s-1 in still water. A floating log is at 80 m straight across and drifts downstream at 6 m s-1. Find the least time to reach the log.

Key Formula: In water frame, the log is stationary across the river; least time = across separation / boat speed.

  1. Transform to the water frame moving with the current.
  2. The log has no velocity in the water frame and remains 80 m across from the boat's initial line.
  3. The boat should head directly toward the log's water-frame position, so time = 80/10 = 8 s.

Answer: 8 s.

JEE Advanced Difficult Questions

These are JEE Advanced style original questions. They are not claimed as previous-year problems.

DifficultJEE Advanced style original question

JEE Advanced Style Original 1: Multiple-correct river crossing limits

Question: A boat has speed u relative to water and the river current has speed v, with u > v. River width is d. The boatman can choose any fixed heading. Which statements are correct?

Key Concept Tested: Velocity triangle, shortest time, shortest path

Complete Solution: Minimum time is d/u when the boat is aimed normal to the bank. For reaching the exactly opposite point, the upstream component must cancel v, so the across speed becomes √(u2 - v2) and the time is d/√(u2 - v2). The no-drift path is longer in time unless v = 0.

Final Answer: Correct statements: shortest time is d/u; shortest path time is d/√(u2 - v2); shortest path requires an upstream heading.

DifficultJEE Advanced style original question

JEE Advanced Style Original 2: Interception of a drifting target

Question: A swimmer can swim at speed u in still water. A floating target is initially at vector R = a i + d j relative to the swimmer, where i is downstream and j is across the river. The current is v i. Find the minimum interception time if the swimmer may choose heading freely.

Key Concept Tested: Relative motion in the water frame

Complete Solution: Shift to the water frame. Both swimmer and target share the same current v i, so the target is stationary at R in that frame. The shortest interception path is the straight segment of length |R| at speed u.

Final Answer: tmin = √(a2 + d2)/u.

DifficultJEE Advanced style original question

JEE Advanced Style Original 3: Closest approach using dot product

Question: At t = 0, particle B is at 30 i + 40 j m relative to A. Their relative velocity vBA is -6 i - 8 j m s-1. Determine whether they collide and, if not, the closest distance.

Key Concept Tested: Relative position line and collision condition

Complete Solution: The initial relative position is r = 30 i + 40 j. Since vBA = -0.2(30 i + 40 j), the relative velocity is exactly opposite to r. The line of relative motion passes through the origin. Time to collision is |r|/|v| = 50/10 = 5 s.

Final Answer: They collide after 5 s; closest distance is 0.

DifficultJEE Advanced style original question

JEE Advanced Style Original 4: Rain observed by two opposite runners

Question: Two runners move east and west with equal speed u. The eastward runner sees rain making angle α with vertical toward his front, while the westward runner sees angle β with vertical toward his front. Express the horizontal rain speed in terms of u, α, and β.

Key Concept Tested: Solving actual rain velocity from two relative observations

Complete Solution: Let actual horizontal rain velocity be w east and vertical downward speed be V. For the eastward runner, tan α = (u - w)/V if rain appears from his front. For the westward runner, tan β = (u + w)/V. Adding gives V = 2u/(tan α + tan β). Subtracting gives w = V(tan β - tan α)/2.

Final Answer: w = u(tan β - tan α)/(tan α + tan β), east positive.

DifficultJEE Advanced style original question

JEE Advanced Style Original 5: Aircraft with fuel-time constraint

Question: An aircraft of airspeed u must fly due north over the ground in a steady east wind v, with u > v. Fuel lasts for time T. Find the maximum northward ground distance.

Key Concept Tested: Crosswind correction and useful component

Complete Solution: The aircraft must aim west of north so that its westward component equals v. The remaining northward component is √(u2 - v2). During time T, ground displacement along north is this component times T.

Final Answer: Dmax = T√(u2 - v2).

DifficultJEE Advanced style original question

JEE Advanced Style Original 6: Locus of relative position

Question: The position of B relative to A is rBA = (12 - 3t)i + (5 + 4t)j in metres. Find the locus and the minimum separation.

Key Concept Tested: Parametric relative motion and closest approach

Complete Solution: The relative path is a straight line because each component is linear in t. Initial r = 12 i + 5 j and relative velocity v = -3 i + 4 j. Minimum distance squared is |r|2 - (r.v)2/|v|2. Here |r|2 = 169, r.v = -36 + 20 = -16, |v|2 = 25. Thus dmin2 = 169 - 256/25 = 3969/25.

Final Answer: The locus is a straight line; minimum separation = 63/5 m.

DifficultJEE Advanced style original question

JEE Advanced Style Original 7: Matching river quantities

Question: Match the river-crossing objective with the governing expression for a boat speed u, current v, river width d, and u > v: shortest time, no drift, drift when aimed normally, and no-drift heading.

Key Concept Tested: Recognizing river-case formulas

Complete Solution: Shortest time uses full across speed: d/u. No drift has across speed √(u2 - v2). Aimed normal gives time d/u and drift vd/u. The no-drift heading satisfies u sin θ = v where θ is upstream from the normal.

Final Answer: Shortest time -> d/u; no-drift time -> d/√(u2 - v2); normal-heading drift -> vd/u; heading -> sin θ = v/u.

DifficultJEE Advanced style original question

JEE Advanced Style Original 8: Landing at a prescribed downstream point

Question: A boat of speed u crosses a river of width d and current v. It must land at a point a downstream from the point exactly opposite. Derive the time equation for a straight heading.

Key Concept Tested: Combining prescribed displacement with velocity constraints

Complete Solution: Ground velocity components must be a/t downstream and d/t across. Boat velocity relative to water is ground velocity minus current: (a/t - v)i + (d/t)j. Its magnitude must equal u. Therefore (a/t - v)2 + (d/t)2 = u2. Multiplying by t2 gives a quadratic in t.

Final Answer: (u2 - v2)t2 + 2avt - (a2 + d2) = 0; take the positive root.

DifficultJEE Advanced style original question

JEE Advanced Style Original 9: Graph interpretation of relative motion

Question: A relative position graph rBA(t) is a straight line in the xy-plane passing closest to the origin at t = 3 s. What physical condition holds at t = 3 s?

Key Concept Tested: Geometric meaning of closest approach

Complete Solution: Closest approach occurs when the relative position vector is perpendicular to the relative velocity vector. On a straight-line relative path, the perpendicular from the origin to the line marks the nearest point.

Final Answer: rBA(3).vBA = 0.

DifficultJEE Advanced style original question

JEE Advanced Style Original 10: Collision condition in two dimensions

Question: Two particles have initial relative position r and constant relative velocity v. State the necessary and sufficient condition for collision at a future time.

Key Concept Tested: Vector collinearity plus positive time

Complete Solution: For collision, r + vt = 0 must have a single positive t satisfying both components. Thus r and v must be anti-parallel, and the ratio of corresponding components must give t > 0. Equivalently, r x v = 0 and r.v < 0.

Final Answer: Collision occurs iff r x v = 0 and r.v < 0; then t = |r|/|v|.

DifficultJEE Advanced style original question

JEE Advanced Style Original 11: Minimum-time chase with moving target

Question: A pursuer at the origin moves with speed u in still air. A target is initially at R and moves with constant velocity w. Derive the equation for interception time.

Key Concept Tested: Relative displacement with chosen pursuer direction

Complete Solution: At interception, the pursuer displacement has magnitude ut and equals R + wt. Therefore |R + wt| = ut. Squaring gives (w.w - u2)t2 + 2(R.w)t + R.R = 0. A positive root gives the possible interception time.

Final Answer: (|w|2 - u2)t2 + 2(R.w)t + |R|2 = 0.

DifficultJEE Advanced style original question

JEE Advanced Style Original 12: Maximum allowed drift

Question: A boat of speed u crosses a river of width d and current v. It must keep downstream drift not greater than x. Write the condition on crossing time t for a feasible straight-line heading.

Key Concept Tested: Feasibility under displacement constraint

Complete Solution: If drift is x or less, ground velocity components are at most x/t downstream and exactly d/t across. The boat velocity relative to water is (x/t - v)i + (d/t)j for boundary drift. Feasibility requires a heading whose required boat-speed magnitude is no more than u.

Final Answer: For the boundary x, feasibility requires (x/t - v)2 + (d/t)2 ≤ u2; some t > 0 satisfying this gives an allowed path.

DifficultJEE Advanced style original question

JEE Advanced Style Original 13: Multiple-correct rain-frame statements

Question: Rain velocity relative to ground is R and observer velocity is U. Which statements are always correct: apparent rain velocity is R - U; umbrella direction is along apparent rain; increasing observer speed can change apparent angle; apparent vertical rain implies R has no horizontal component.

Key Concept Tested: Relative velocity and apparent direction

Complete Solution: The first three statements follow directly from relative velocity. The last statement is false because apparent vertical rain requires Rx - Ux = 0, so R may have a horizontal component equal to the observer's horizontal velocity.

Final Answer: Correct: first, second, and third statements only.

DifficultJEE Advanced style original question

JEE Advanced Style Original 14: Shortest path at limiting boat speed

Question: A boat speed u equals current speed v. Discuss whether a finite-time no-drift crossing is possible.

Key Concept Tested: Limiting case of velocity triangle

Complete Solution: No drift requires the upstream component of boat velocity to equal v. If u = v, the entire boat speed must be upstream, leaving zero across component. With zero across velocity, a finite river width cannot be crossed in finite time.

Final Answer: No finite-time no-drift crossing is possible; the required crossing time tends to infinity.

DifficultJEE Advanced style original question

JEE Advanced Style Original 15: Paragraph-based two-observer velocity reconstruction

Question: Observer P moves east at 3 m s-1 and sees a drone moving north at 4 m s-1 relative to him. Observer Q is at rest on ground. Find the drone velocity seen by Q and its speed.

Key Concept Tested: Adding observer velocity back to relative velocity

Complete Solution: Velocity of drone relative to P is 0 i + 4 j. Since P moves at 3 i relative to ground, drone velocity relative to ground is vD = vDP + vP = 3 i + 4 j.

Final Answer: 3 i + 4 j m s-1; speed = 5 m s-1.

IB, ICSE, IGCSE and A-Level Physics Questions

Structured questions for different curricula, moving from explanation and diagrams to more formal vector calculations.

IB Physics Questions

IB 1: Explaining independence of components

Question: A trolley moves east with constant speed while accelerating north. Explain why x and y components can be analysed separately.

Answer / Explanation: Vector equations split into perpendicular components. A zero east acceleration keeps east velocity constant, while north acceleration changes only the north component. The resultant path is found after combining both components.

IB 2: Drawing a velocity triangle

Question: A boat velocity relative to water is across the river and the current is downstream. Describe the correct vector diagram for ground velocity.

Answer / Explanation: Draw the boat velocity from the tail, then add the current vector downstream from its head. The resultant from the original tail to final head is the boat's velocity relative to ground.

IB 3: Data-based river calculation

Question: A 75 m wide river has current 1.5 m s-1. A boat aimed straight across moves at 2.5 m s-1 relative to water. Calculate time and drift.

Answer / Explanation: Time = 75/2.5 = 30 s. Drift = 1.5 x 30 = 45 m downstream. The current does not change the across component in this case.

IB 4: Assumptions in rain-man problems

Question: State two assumptions often made when solving umbrella-direction problems.

Answer / Explanation: Rain velocity is treated as uniform over the small region considered, and the observer moves with constant velocity. Air turbulence, changing drop size, and gusts are neglected.

IB 5: Relative velocity and frame choice

Question: Why can a floating log be treated as stationary in the water frame?

Answer / Explanation: A floating log moves with the current. In a frame moving with the current, the log's current velocity is removed, making it stationary unless it has its own propulsion.

IB 6: Uncertainty in component measurements

Question: Velocity components are measured as 6.0 +/- 0.2 m s-1 and 8.0 +/- 0.2 m s-1. Explain why the speed uncertainty is not found by simply adding speeds.

Answer / Explanation: The speed is calculated using a square root of squared components. Uncertainties must be propagated through the function v = √(vx2 + vy2), so component contributions are weighted by their fractions of the resultant.

IB 7: Real-life crosswind application

Question: Explain why a pilot may point the aircraft slightly west of north while the ground track is exactly north.

Answer / Explanation: If wind pushes the aircraft east, the aircraft must have a westward air-velocity component. The vector sum of air velocity and wind then has no east-west component, leaving a northward ground velocity.

IB 8: Speed versus velocity in circular-looking paths

Question: A student says a constant speed means constant velocity. Use two-dimensional motion to correct the statement.

Answer / Explanation: Velocity includes direction. In two-dimensional motion, an object can keep the same speed while changing direction, so its velocity changes and it has acceleration.

IB 9: Interpreting relative motion of two cyclists

Question: Two cyclists ride with the same velocity side by side. What does each cyclist observe about the other, and why?

Answer / Explanation: Each observes the other at rest because their relative velocity is zero. Their positions may differ, but their separation vector remains constant.

IB 10: Evaluating model limitations

Question: A river-current calculation predicts an exact landing point. Give one reason the real boat might miss it.

Answer / Explanation: The model usually assumes steady current and constant boat speed. Real current can vary across the river, steering may change, and wind can add another velocity component.

ICSE Physics Questions

ICSE 1: Meaning of scalar speed

Question: What is the difference between speed and velocity?

Answer / Explanation: Speed has magnitude only. Velocity has magnitude and direction, so it is a vector.

ICSE 2: Simple resultant velocity

Question: A person walks east at 3 m s-1 and north at 4 m s-1 as components of motion. Find the resultant speed.

Answer / Explanation: Resultant speed = √(32 + 42) = 5 m s-1.

ICSE 3: Position vector definition

Question: Write the position vector of a point (2 m, 5 m) from the origin.

Answer / Explanation: r = 2 i + 5 j m.

ICSE 4: Relative speed in same direction

Question: Two cars move in the same direction at 18 m s-1 and 12 m s-1. Find relative speed.

Answer / Explanation: Relative speed = 18 - 12 = 6 m s-1.

ICSE 5: Relative speed in opposite directions

Question: Two bicycles move towards each other at 5 m s-1 and 7 m s-1. Find relative speed.

Answer / Explanation: Relative speed = 5 + 7 = 12 m s-1.

ICSE 6: River current effect

Question: If a boat is aimed straight across a river, what does the current do?

Answer / Explanation: The current carries the boat downstream while it crosses. It causes drift but does not by itself reduce the boat's across component.

ICSE 7: Umbrella direction

Question: Why does a walking person tilt an umbrella forward in vertical rain?

Answer / Explanation: Because relative to the walking person, the rain has a backward horizontal component. The umbrella is tilted along the apparent rain direction.

ICSE 8: Acceleration as vector

Question: Can acceleration be in a different direction from velocity?

Answer / Explanation: Yes. Acceleration tells how velocity changes; it may change speed, direction, or both.

ICSE 9: Unit vector notation

Question: What do i and j represent in r = x i + y j?

Answer / Explanation: They represent unit vectors along the x-axis and y-axis respectively.

ICSE 10: No relative motion

Question: When will two moving objects appear at rest relative to each other?

Answer / Explanation: They appear at rest relative to each other when they have the same velocity vector.

IGCSE Physics Questions

IGCSE 1: Resultant of perpendicular velocities

Question: A boat moves 4 m s-1 across a river while the current is 3 m s-1 downstream. Find resultant speed.

Answer / Explanation: Resultant speed = √(42 + 32) = 5 m s-1.

IGCSE 2: Vector diagram for wind

Question: How should you draw a wind vector diagram for an aircraft?

Answer / Explanation: Draw the aircraft velocity relative to air and add the wind velocity tip-to-tail. The resultant is the aircraft velocity relative to ground.

IGCSE 3: Speed and direction

Question: Why is 20 m s-1 east a velocity but 20 m s-1 is only a speed?

Answer / Explanation: The first includes direction, so it is a vector velocity. The second gives only magnitude.

IGCSE 4: Relative motion on a train

Question: A passenger walks forward at 2 m s-1 inside a train moving at 18 m s-1. Find passenger speed relative to ground.

Answer / Explanation: If the walking direction is the same as the train, speed relative to ground = 18 + 2 = 20 m s-1.

IGCSE 5: Components of displacement

Question: A student walks 6 m east and 8 m north. Find the straight-line displacement.

Answer / Explanation: Displacement magnitude = √(62 + 82) = 10 m.

IGCSE 6: Rain and moving observer

Question: Rain falls vertically. A cyclist moves forward. In which direction does the rain appear to come from?

Answer / Explanation: It appears to come from ahead of the cyclist because the cyclist's forward motion creates a backward relative component of rain.

IGCSE 7: Boat drift

Question: A boat takes 20 s to cross a river with current 1.2 m s-1. How far downstream does it drift?

Answer / Explanation: Drift = 1.2 x 20 = 24 m.

IGCSE 8: Direction from components

Question: A velocity has components 0 m s-1 east and 5 m s-1 north. What is its direction?

Answer / Explanation: The velocity is directed due north because the east component is zero.

IGCSE 9: Relative speed of walkers

Question: Two walkers move in opposite directions at 1.5 m s-1 and 2.0 m s-1. Find their relative speed.

Answer / Explanation: Relative speed = 1.5 + 2.0 = 3.5 m s-1.

IGCSE 10: Adding vector components

Question: Why can perpendicular vector components be added using Pythagoras?

Answer / Explanation: They form a right-angled triangle, so the resultant is the hypotenuse and its magnitude follows Pythagoras' theorem.

A-Level Physics Questions

A-Level 1: Bearing and component velocity

Question: A boat travels at 12 m s-1 on a bearing 060°. Find east and north components.

Answer / Explanation: Bearing is measured clockwise from north. East component = 12 sin 60° = 6√3 m s-1; north component = 12 cos 60° = 6 m s-1.

A-Level 2: Crosswind correction calculation

Question: An aircraft airspeed is 90 m s-1. Wind is 18 m s-1 from west to east. Find heading correction for a north ground track.

Answer / Explanation: The aircraft needs an 18 m s-1 westward component. sin θ = 18/90 = 0.2, so it heads θ = sin-1(0.2) west of north.

A-Level 3: Closest distance between moving objects

Question: Relative position is r = 10 i + 24 j m and relative velocity is v = 3 i - 4 j m s-1. Find time of closest approach.

Answer / Explanation: t = -(r.v)/|v|2. Here r.v = 30 - 96 = -66 and |v|2 = 25, so t = 66/25 s.

A-Level 4: River crossing with given drift

Question: A boat crosses a 160 m river in 40 s and drifts 60 m downstream. Current is 2 m s-1. Find boat velocity relative to water.

Answer / Explanation: Ground components are 4 m s-1 across and 1.5 m s-1 downstream. Subtract current: boat relative to water has 4 m s-1 across and 0.5 m s-1 upstream.

A-Level 5: Parametric position and acceleration

Question: A particle has r = (t2 + 2t)i + (3t - 1)j m. Find velocity and acceleration.

Answer / Explanation: v = dr/dt = (2t + 2)i + 3j m s-1. a = dv/dt = 2i m s-2.

A-Level 6: Resultant velocity from two angled vectors

Question: Two velocity components of magnitudes 10 m s-1 and 6 m s-1 act at 60°. Find resultant magnitude.

Answer / Explanation: Use cosine rule: R = √(102 + 62 + 2(10)(6)cos60°) = √196 = 14 m s-1.

A-Level 7: Relative acceleration

Question: If aA = 5 i - 2 j and aB = -i + 4 j m s-2, find acceleration of A relative to B.

Answer / Explanation: aAB = aA - aB = (5 - (-1))i + (-2 - 4)j = 6 i - 6 j m s-2.

A-Level 8: Ground velocity from air velocity and wind

Question: A drone has air velocity 15 i + 20 j m s-1. Wind velocity is -5 i + 2 j m s-1. Find ground velocity and speed.

Answer / Explanation: Ground velocity = (15 - 5)i + (20 + 2)j = 10 i + 22 j m s-1. Speed = √(102 + 222) = 2√146 m s-1.

A-Level 9: Direction of no-drift boat heading

Question: A boat speed is 17 m s-1 and current is 8 m s-1. Find the angle upstream from the normal for no drift.

Answer / Explanation: No drift requires 17 sin θ = 8. Therefore θ = sin-1(8/17) upstream from the normal.

A-Level 10: Interpreting observer frame transformation

Question: An object velocity in frame S is 9 i + 12 j m s-1. Frame S' moves at 4 i m s-1 relative to S. Find object velocity in S'.

Answer / Explanation: Subtract the frame velocity: v' = (9 i + 12 j) - 4 i = 5 i + 12 j m s-1.

Assertion-Reason Questions

These 10 original assertion-reason items target common conceptual traps in relative velocity, river crossing, umbrella direction and wind correction.

Assertion-Reason 1: Relative velocity formula

Assertion: Velocity of A with respect to B is vA - vB.

Reason: Relative velocity is found by subtracting the observer's velocity from the object's velocity.

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Assertion-Reason 2: Shortest river-crossing time

Assertion: For shortest time across a river, the boat should be aimed perpendicular to the bank.

Reason: This makes the across-river component of boat velocity maximum.

Answer: Both are true. The shortest-time condition is controlled only by the across-river component, so maximizing that component gives the least time.

Assertion-Reason 3: Shortest river path

Assertion: For shortest path across a river, the resultant ground velocity must be perpendicular to the bank.

Reason: Shortest distance between opposite banks is the straight line normal to the banks.

Answer: Both are true. A perpendicular ground velocity removes downstream drift, making the track the shortest line between banks.

Assertion-Reason 4: Umbrella direction

Assertion: A walking person tilts an umbrella even when rain falls vertically in the ground frame.

Reason: The rain has a horizontal component relative to the walking person.

Answer: Both are true. The umbrella is aligned with apparent rain, and the apparent horizontal component appears because the observer is moving.

Assertion-Reason 5: No-drift possibility

Assertion: A boat slower than the river current cannot reach the exactly opposite point by a straight no-drift crossing.

Reason: Its maximum upstream component is less than the current speed.

Answer: Both are true. No drift needs full cancellation of current, which is impossible when the boat cannot supply enough upstream component.

Assertion-Reason 6: Tailwind effect

Assertion: A tailwind increases an aircraft's ground speed.

Reason: Tailwind velocity is added in the same direction as aircraft air velocity.

Answer: Both are true. Since the wind vector points along the aircraft's motion, the ground-speed magnitude becomes larger than the airspeed.

Assertion-Reason 7: Headwind travel time

Assertion: For a fixed distance, a headwind increases travel time.

Reason: A headwind decreases ground speed below airspeed.

Answer: Both are true. With smaller ground speed, the same distance requires a longer time by t = distance/speed.

Assertion-Reason 8: Observer-dependent rest

Assertion: A moving train passenger may see another passenger at rest.

Reason: Two objects with equal velocity vectors have zero relative velocity.

Answer: Both are true. If both passengers share the same velocity, their relative position stays fixed inside the train frame.

Assertion-Reason 9: Current and boat speed

Assertion: River current always changes the boat's speed relative to water.

Reason: Current changes the boat's velocity relative to ground.

Answer: Assertion is false, but Reason is true. Current affects ground velocity, not the boat's speed relative to water.

Assertion-Reason 10: Crosswind correction

Assertion: In crosswind, an aircraft may point in a direction different from its ground track.

Reason: Ground velocity is the vector sum of air velocity and wind velocity.

Answer: Both are true. The heading is chosen so the wind-corrected vector sum lies along the required ground track.

Case Study Questions

Each case study connects formulas to a physical situation and includes direct answers with reasoning.

Case Study 1: Boat crossing a river

Passage: A boat can move at 5 m s-1 relative to water. The river is 100 m wide and flows at 3 m s-1 downstream.

  1. What is the shortest crossing time?
    20 s, because the boat must use its full 5 m s-1 across the river.
  2. What is the drift in shortest-time crossing?
    Drift = 3 x 20 = 60 m downstream.
  3. Can the boat reach exactly opposite?
    Yes, because boat speed 5 m s-1 is greater than current 3 m s-1.
  4. What is the no-drift crossing time?
    Across component = √(52 - 32) = 4 m s-1; time = 100/4 = 25 s.

Case Study 2: Rain and moving observer

Passage: Rain falls vertically downward at 15 m s-1. A cyclist moves east at 8 m s-1.

  1. What is the relative rain velocity?
    vrain,cycle = -8 i - 15 j m s-1, taking east as +i and upward as +j.
  2. What angle does apparent rain make with the vertical?
    tan θ = 8/15.
  3. Which way should the cyclist tilt the umbrella?
    Forward, toward the direction of motion, along the apparent rain direction.
  4. Does the actual rain have a horizontal component?
    No. The horizontal component appears only in the cyclist's frame.

Case Study 3: Airplane in crosswind

Passage: A plane has airspeed 300 km h-1. A wind blows east at 60 km h-1. The required ground track is due north.

  1. Which side of north should the plane head?
    West of north, so that its westward component cancels the east wind.
  2. What is the heading condition?
    300 sin θ = 60, so sin θ = 0.2.
  3. What is the northward ground speed?
    √(3002 - 602) = 120√6 km h-1.
  4. Why is this a vector addition problem?
    The ground velocity equals aircraft velocity relative to air plus wind velocity.

Case Study 4: Two particles moving in 2D

Passage: Particle A is at the origin and moves with velocity 4 i + 3 j m s-1. Particle B is initially at 20 i m and moves with velocity -1 i + 3 j m s-1.

  1. What is velocity of B relative to A?
    vBA = (-1 i + 3 j) - (4 i + 3 j) = -5 i m s-1.
  2. Do they ever have the same y-coordinate?
    They always have the same relative y-velocity, so if their initial y-coordinates are equal, they stay equal in y.
  3. When do they meet?
    Initial x-separation is 20 m and relative x-speed is 5 m s-1, so they meet after 4 s.
  4. What concept is tested?
    Collision through relative position becoming zero.

Case Study 5: Rescue interception problem

Passage: A rescue boat at O can move at 12 m s-1 relative to water. A person is floating 90 m across the river and 40 m downstream from O. The current carries both downstream at 5 m s-1.

  1. Which frame makes the problem easiest?
    The water frame, because the floating person is at rest in that frame.
  2. What distance must the boat cover in the water frame?
    √(902 + 402) = 10√97 m.
  3. What is the minimum rescue time?
    (10√97)/12 = 5√97/6 s.
  4. What happens in the ground frame during this time?
    Both the person and boat are additionally carried downstream by the current, but interception time is unchanged.

Need Personal Help With Relative Velocity?

If Motion in a Plane, Relative Velocity, River-Boat Problems, Rain-Man Problems, Airplane Wind Problems, Vector Components or JEE Advanced level relative motion questions are still not clear, you can directly contact Kumar Sir for one-to-one online Physics classes.

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