Position-Time Graph
Slope of x-t graph gives velocity: v = Δx/Δt. Steeper slope means greater speed. Horizontal line means rest.
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Motion Graphs Formula Sheet
Velocity-Time Graph
Slope of v-t graph gives acceleration: a = Δv/Δt. Area under v-t graph gives displacement.
Acceleration-Time Graph
Area under a-t graph gives change in velocity: Δv = area under a-t graph.
Average Velocity
Average velocity = total displacement / total time. It can be positive, negative or zero.
Average Speed
Average speed = total path length / total time. It is always non-negative.
Instantaneous Velocity
Instantaneous velocity is slope of tangent on x-t graph at that instant.
Instantaneous Acceleration
Instantaneous acceleration is slope of tangent on v-t graph at that instant.
Graph Shortcuts
x-t slope → v, v-t slope → a, v-t area → displacement, a-t area → change in velocity.
NCERT Solutions: Exercises 2.1 to 2.18
Click any question to view detailed solution, final answer and exam tip.
2.1In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Detailed Solution
A body is treated as a point object when its size is negligible compared with the distance travelled and rotation or internal structure is not important. In (a), the railway carriage travels a distance much larger than its size, so it may be treated as a point object. In (b), the monkey-man system moves along a circular track, and if the track radius is large compared with the body size, it may be approximately a point object. In (c), spin and sharp turning are essential, so finite size and rotation matter. In (d), tumbling means rotation and size are important.
Final Answer
(a) and (b), approximately.
Exam Tip: For point object questions, compare object size with distance travelled and check whether rotation matters.
2.2The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below;
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice).
Detailed Solution
In an x-t graph, distance from origin is shown on vertical axis and time on horizontal axis. P is below Q, so P is closer to school than Q. A reaches P and B reaches Q; hence A lives closer than B. The line for A starts at t=0, while B starts later, so A starts earlier. Speed is slope of x-t graph. B has a steeper line, so B walks faster. Both lines end at the same vertical time line, so they reach home at the same time. The two lines intersect once, so B overtakes A once.
Final Answer
(a) A lives closer than B. (b) A starts earlier than B. (c) B walks faster than A. (d) same time. (e) B overtakes A once.
Exam Tip: On x-t graphs, steeper slope means greater speed.
2.3A woman starts from her home at 9.00 am, walks with a speed of 5 km h-1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h-1. Choose suitable scales and plot the x-t graph of her motion.
Detailed Solution
Time to reach office = distance/speed = 2.5/5 = 0.5 h = 30 min. So she reaches office at 9:30 am. She stays at office until 5:00 pm, so position remains x=2.5 km for 7.5 h. Return time = 2.5/25 = 0.1 h = 6 min. Thus she reaches home at 5:06 pm. The x-t graph is a straight rising line from (9:00,0) to (9:30,2.5), a horizontal line till (5:00,2.5), then a steep falling line to (5:06,0).
Final Answer
Graph: rising straight line for 30 min, horizontal line from 9:30 am to 5:00 pm, steep falling line from 5:00 pm to 5:06 pm.
Exam Tip: Horizontal x-t graph means rest; steeper line means larger speed.
2.4A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Detailed Solution
In each cycle of 8 steps, net displacement = 5 m - 3 m = 2 m and time = 8 s. After 4 cycles, displacement = 8 m and time = 32 s. Now he moves forward: 9 m at 33 s, 10 m at 34 s, 11 m at 35 s, 12 m at 36 s, and 13 m at 37 s. So he falls into the pit at 37 s, before stepping backward in that cycle. The x-t graph is a zig-zag: five rising unit steps followed by three falling unit steps, repeated.
Final Answer
Time taken = 37 s.
Exam Tip: Do not use only average speed for the last cycle; check when the pit is first reached.
2.5A car moving along a straight highway with speed of 126 km h-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Detailed Solution
Initial speed u = 126 km h-1 = 126 × 5/18 = 35 m s-1. Final speed v=0 and s=200 m. From v2=u2+2as, 0=352+2a(200). Hence a=-1225/400=-3.0625 m s-2. Retardation magnitude = 3.06 m s-2. Time from v=u+at gives 0=35-3.0625t, so t=11.43 s.
Final Answer
Retardation = 3.06 m s-2; stopping time = 11.43 s.
Exam Tip: Convert km h-1 to m s-1 before using equations.
2.6A player throws a ball upwards with an initial speed of 29.4 m s-1.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of the x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s-2 and neglect air resistance).
Detailed Solution
(a) Acceleration due to gravity is always vertically downward. (b) At the highest point, velocity is zero but acceleration remains downward with magnitude g. (c) If downward is positive and origin is at the highest point, the ball is below the origin during both upward and downward parts, so position x is positive. During upward motion velocity is upward, opposite to positive direction, so velocity is negative. During downward motion velocity is positive. Acceleration is downward, so acceleration is positive throughout. (d) Maximum height H = u2/2g = 29.42/(2×9.8)=44.1 m. Time of ascent = u/g = 29.4/9.8=3 s. Total time of flight = 6 s.
Final Answer
(a) Downward. (b) v=0, a=9.8 m s-2 downward. (c) Upward: x positive, v negative, a positive; downward: x positive, v positive, a positive. (d) Height = 44.1 m; total time = 6 s.
Exam Tip: At the top velocity is zero, acceleration is not zero.
2.7Read each statement below carefully and state with reasons and examples, if it is true or false:
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Detailed Solution
(a) True. At the highest point of vertical motion, speed is zero but acceleration is g downward. (b) False. Speed is the magnitude of velocity, so zero speed means velocity is zero. (c) True for one-dimensional motion if speed is constant and direction cannot change without passing through zero; velocity remains constant, so acceleration is zero. (d) False. If velocity is negative and acceleration is positive, the magnitude of velocity decreases, so the particle slows down.
Final Answer
(a) True, (b) False, (c) True in one-dimensional motion, (d) False.
Exam Tip: Positive acceleration does not always mean speeding up; compare signs of velocity and acceleration.
2.8A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Detailed Solution
For the first fall, u=0 and h=90 m. With g=10 m s-2, h=½gt2 gives 90=5t2, so t≈4.24 s. Just before hitting the floor, speed v=gt≈42.4 m s-1. After collision it loses one tenth of speed, so rebound speed becomes 0.9v≈38.2 m s-1. During upward motion, speed decreases linearly to zero, then increases linearly while falling back. Each collision produces a sudden drop in speed to 0.9 of the previous impact speed. Hence the speed-time graph consists of straight-line segments with sudden vertical drops at collisions.
Final Answer
Speed-time graph: repeated straight-line V-shaped segments with vertical drops at each bounce; first impact at about 4.24 s, and each rebound speed is 0.9 times the preceding impact speed.
Exam Tip: In speed-time graphs, speed is never negative; bouncing causes sudden discontinuities.
2.9Explain clearly, with examples, the distinction between:
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only].
Detailed Solution
Magnitude of displacement is the shortest separation between initial and final positions, while total path length is the actual distance travelled along the path. In one dimension, if a particle reverses direction, path length becomes larger than displacement magnitude. Average velocity magnitude = |displacement|/time. Average speed = total path length/time. Since total path length is always greater than or equal to displacement magnitude, average speed is greater than or equal to magnitude of average velocity. Equality holds only when the particle moves in one direction without reversing.
Final Answer
Total path length ≥ magnitude of displacement. Average speed ≥ magnitude of average velocity. Equality holds when there is no change in direction.
Exam Tip: For return journeys, displacement may be zero but path length is not zero.
2.10A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h-1. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]
Detailed Solution
Speed to market = 5 km h-1. Time to market = 2.5/5=0.5 h=30 min. Return speed = 7.5 km h-1. Time to return = 2.5/7.5=1/3 h=20 min. (i) In 0 to 30 min, displacement=2.5 km, distance=2.5 km, time=0.5 h. Average velocity magnitude=5 km h-1, average speed=5 km h-1. (ii) In 0 to 50 min, he is back home. Displacement=0, total distance=5 km, time=50 min=5/6 h. Average velocity magnitude=0; average speed=5/(5/6)=6 km h-1. (iii) In 0 to 40 min, he goes to market in 30 min and returns for 10 min. Return distance in 10 min = 7.5×(1/6)=1.25 km. Position from home = 2.5-1.25=1.25 km. Total distance=2.5+1.25=3.75 km, time=2/3 h. Average velocity magnitude=1.25/(2/3)=1.875 km h-1. Average speed=3.75/(2/3)=5.625 km h-1.
Final Answer
(i) 5 km h-1, 5 km h-1. (ii) 0, 6 km h-1. (iii) 1.875 km h-1, 5.625 km h-1.
Exam Tip: Average speed uses total distance; average velocity uses displacement.
2.11In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Detailed Solution
Instantaneous velocity is defined over an infinitesimally small time interval. In such a very small interval, the particle cannot execute a finite reversal of direction; the path length and displacement magnitude become the same in the limiting sense. Therefore instantaneous speed, which is rate of change of path length, equals the magnitude of instantaneous velocity.
Final Answer
Instantaneous speed = magnitude of instantaneous velocity because over an infinitesimal interval distance and displacement magnitude are equal.
Exam Tip: The distinction matters for average quantities, not instantaneous quantities.
2.12Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
Detailed Solution
In one-dimensional motion, a particle cannot have two positions at the same instant. Therefore an x-t graph must not loop back vertically in a way that gives more than one x for the same t. Graph (a) fails this condition. A v-t graph may have positive and negative velocities at different times, but not two velocities at the same instant. Graph (b) is a closed loop in v-t plane, giving multiple velocities for the same time; impossible. Graph (c), speed-time graph, goes below the time axis in the screenshot style only if interpreted as speed becoming negative; speed cannot be negative, so any negative part is impossible. Graph (d) represents total path length versus time decreasing at parts; total path length cannot decrease.
Final Answer
Graphs (a), (b), (c) if speed is negative, and (d) cannot represent valid one-dimensional motion as shown.
Exam Tip: Check whether one time gives more than one value, and remember speed/path length cannot be negative or decrease.
2.13Figure 2.11 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.
Detailed Solution
No. The graph is x versus t, not the actual path in space. Since the motion is one-dimensional, the particle always moves along a straight line. The curve in x-t graph tells how its position changes with time. For t<0, x is constant, so the particle is at rest. For t>0, x increases non-linearly with time, which may represent accelerated motion along a straight line, such as a car starting from rest and speeding up.
Final Answer
It is not a spatial path. Suitable context: a particle at rest before t=0, then moving along a straight line with increasing velocity after t=0.
Exam Tip: Never confuse x-t graph shape with the actual trajectory.
2.14A police van moving on a highway with a speed of 30 km h-1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h-1. If the muzzle speed of the bullet is 150 m s-1, with what speed does the bullet hit the thief’s car? (Note: Obtain that speed which is relevant for damaging the thief’s car).
Detailed Solution
Convert speeds: police van speed = 30 km h-1 = 8.33 m s-1. Thief car speed = 192 km h-1 = 53.33 m s-1. Bullet speed relative to ground = muzzle speed relative to police van + police van speed = 150 + 8.33 = 158.33 m s-1. The damaging speed is relative speed of bullet with respect to thief’s car = 158.33 - 53.33 = 105 m s-1.
Final Answer
Bullet hits the thief’s car with relative speed 105 m s-1.
Exam Tip: Damage depends on relative speed, not ground speed.
2.15Suggest a suitable physical situation for each of the following graphs (Fig 2.12):
Detailed Solution
(a) The x-t graph first shows rest at a fixed position, then motion in positive direction, then motion in negative direction crossing the original level, and finally rest at another position. A suitable example is a lift going up, then coming down below starting level, and stopping. (b) The v-t graph shows repeated straight segments with negative slope separated by jumps. This may describe a ball thrown upward repeatedly, or a bouncing ball where velocity changes suddenly at impacts. (c) The a-t graph shows acceleration zero except for a short sharp pulse. This may represent a ball hit by a bat, where acceleration is very large for a very small time interval.
Final Answer
(a) Lift or vehicle moving, reversing and stopping. (b) Repeated upward throws or bouncing motion with sudden velocity changes. (c) A short impact such as bat hitting a ball.
Exam Tip: For graph situation questions, translate slope and sign into real motion words.
2.16Figure 2.13 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, -1.2 s.
Detailed Solution
For SHM, acceleration is opposite in sign to displacement: a = -ω2x. Velocity sign is the sign of slope of the x-t graph. From Fig. 2.13, at t=0.3 s the graph is below the time axis and falling, so x is negative, v is negative and a is positive. At t=1.2 s the graph is above the time axis and rising toward a crest, so x is positive, v is positive and a is negative. At t=-1.2 s the graph is below the time axis and rising toward the axis, so x is negative, v is positive and a is positive.
Final Answer
At 0.3 s: x -, v -, a +. At 1.2 s: x +, v +, a -. At -1.2 s: x -, v +, a +.
Exam Tip: In x-t graphs, velocity is slope. For SHM, acceleration has sign opposite to displacement.
2.17Figure 2.14 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.
Detailed Solution
Average speed is total path length divided by time interval. Since the time intervals are equal, compare total distance travelled in each interval. In interval 1 the graph rises moderately, so distance is moderate and average velocity is positive. In interval 2 the graph changes rapidly downward after reaching a turning point, so total distance is largest; average velocity is negative because final position is lower than initial position. In interval 3 the graph first moves downward then upward after a minimum, so path length is significant but less than interval 2; average velocity is negative or may be small depending on endpoints shown, but from the figure final position is below initial position in interval 3, so negative. Least average speed is in interval 1.
Final Answer
Greatest average speed: interval 2. Least average speed: interval 1. Average velocity signs: interval 1 positive, interval 2 negative, interval 3 negative.
Exam Tip: Average speed depends on total path length; average velocity sign depends on displacement.
2.18Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ?
Detailed Solution
Average acceleration over an interval equals change in velocity divided by time. Since intervals are equal, compare net change in speed. The greatest magnitude of average acceleration is in the interval where speed changes most between the two ends; from the graph this is interval 2 because speed decreases strongly from near B to C. Average speed is greatest where the area under speed-time graph is greatest; from the graph this is interval 3 because speed remains high and reaches peak D. Since motion is along a constant positive direction, velocity is positive in all intervals. Acceleration sign is slope of speed-time graph: positive where graph rises, negative where graph falls. At A slope is positive, at B slope is zero, at C slope is zero, and at D slope is zero.
Final Answer
Greatest average acceleration magnitude: interval 2. Greatest average speed: interval 3. v is positive in all intervals. Acceleration is positive on rising parts, negative on falling parts. At B, C and D acceleration is zero; at A acceleration is positive.
Exam Tip: For speed-time graph, acceleration is slope and average speed is area/time.
CBSE PYQs and Exam-style Questions
Authentic year is written only when certain; otherwise the card is clearly marked exam-style.
CBSE Exam-style Question 1A straight line x-t graph has positive constant slope. What does it represent?
Solution
Slope of x-t graph is velocity and constant positive slope means constant positive velocity.
Final Answer
Uniform motion in positive direction.
Exam Tip: Slope tells velocity.
CBSE Exam-style Question 2A v-t graph is parallel to time axis above origin. Find acceleration.
Solution
Velocity is constant, so slope of v-t graph is zero.
Final Answer
Zero acceleration.
Exam Tip: Horizontal v-t graph means uniform velocity.
CBSE Exam-style Question 3Area under a v-t graph between two times gives what physical quantity?
Solution
Small area vΔt represents displacement; total area gives displacement.
Final Answer
Displacement.
Exam Tip: Signed area matters.
CBSE Exam-style Question 4Area under an a-t graph gives what physical quantity?
Solution
Since a=Δv/Δt, aΔt gives change in velocity.
Final Answer
Change in velocity.
Exam Tip: Do not confuse with displacement.
CBSE Exam-style Question 5A particle returns to starting point. What is its displacement?
Solution
Initial and final positions are same, so displacement is zero even if distance is non-zero.
Final Answer
Zero.
Exam Tip: Distance and displacement differ.
CBSE Exam-style Question 6For a speed-time graph, what does area represent?
Solution
Speed is non-negative, so area gives total path length.
Final Answer
Distance travelled.
Exam Tip: For speed-time graph, area is distance.
CBSE Exam-style Question 7In an x-t graph, velocity is zero at a turning point. Why?
Solution
Velocity equals slope of x-t graph, and at a smooth maximum or minimum the slope is zero.
Final Answer
Tangent is horizontal.
Exam Tip: Horizontal tangent means zero instantaneous velocity.
CBSE Exam-style Question 8Can acceleration be negative while speed increases?
Solution
If velocity and acceleration have the same sign, speed increases. Thus negative acceleration with negative velocity increases speed.
Final Answer
Yes, if velocity is negative.
Exam Tip: Compare signs of v and a.
CBSE Exam-style Question 9A v-t graph crosses the time axis. What happens to velocity at crossing?
Solution
The sign of velocity changes when graph crosses the time axis.
Final Answer
Velocity is zero and direction may reverse.
Exam Tip: Crossing v=0 often indicates reversal.
CBSE Exam-style Question 10What does a vertical line on an x-t graph imply?
Solution
A vertical line gives many positions at the same time, impossible for one-dimensional motion of one particle.
Final Answer
It is impossible for one particle.
Exam Tip: One time cannot have multiple positions.
CBSE Exam-style Question 11A straight line x-t graph has positive constant slope. What does it represent?
Solution
Slope of x-t graph is velocity and constant positive slope means constant positive velocity.
Final Answer
Uniform motion in positive direction.
Exam Tip: Slope tells velocity.
CBSE Exam-style Question 12A v-t graph is parallel to time axis above origin. Find acceleration.
Solution
Velocity is constant, so slope of v-t graph is zero.
Final Answer
Zero acceleration.
Exam Tip: Horizontal v-t graph means uniform velocity.
NEET PYQs and Exam-style Questions
Authentic year is written only when certain; otherwise the card is clearly marked exam-style.
NEET Exam-style Question 1A straight line x-t graph has positive constant slope. What does it represent?
Solution
Slope of x-t graph is velocity and constant positive slope means constant positive velocity.
Final Answer
Uniform motion in positive direction.
Exam Tip: Slope tells velocity.
NEET Exam-style Question 2A v-t graph is parallel to time axis above origin. Find acceleration.
Solution
Velocity is constant, so slope of v-t graph is zero.
Final Answer
Zero acceleration.
Exam Tip: Horizontal v-t graph means uniform velocity.
NEET Exam-style Question 3Area under a v-t graph between two times gives what physical quantity?
Solution
Small area vΔt represents displacement; total area gives displacement.
Final Answer
Displacement.
Exam Tip: Signed area matters.
NEET Exam-style Question 4Area under an a-t graph gives what physical quantity?
Solution
Since a=Δv/Δt, aΔt gives change in velocity.
Final Answer
Change in velocity.
Exam Tip: Do not confuse with displacement.
NEET Exam-style Question 5A particle returns to starting point. What is its displacement?
Solution
Initial and final positions are same, so displacement is zero even if distance is non-zero.
Final Answer
Zero.
Exam Tip: Distance and displacement differ.
NEET Exam-style Question 6For a speed-time graph, what does area represent?
Solution
Speed is non-negative, so area gives total path length.
Final Answer
Distance travelled.
Exam Tip: For speed-time graph, area is distance.
NEET Exam-style Question 7In an x-t graph, velocity is zero at a turning point. Why?
Solution
Velocity equals slope of x-t graph, and at a smooth maximum or minimum the slope is zero.
Final Answer
Tangent is horizontal.
Exam Tip: Horizontal tangent means zero instantaneous velocity.
NEET Exam-style Question 8Can acceleration be negative while speed increases?
Solution
If velocity and acceleration have the same sign, speed increases. Thus negative acceleration with negative velocity increases speed.
Final Answer
Yes, if velocity is negative.
Exam Tip: Compare signs of v and a.
NEET Exam-style Question 9A v-t graph crosses the time axis. What happens to velocity at crossing?
Solution
The sign of velocity changes when graph crosses the time axis.
Final Answer
Velocity is zero and direction may reverse.
Exam Tip: Crossing v=0 often indicates reversal.
NEET Exam-style Question 10What does a vertical line on an x-t graph imply?
Solution
A vertical line gives many positions at the same time, impossible for one-dimensional motion of one particle.
Final Answer
It is impossible for one particle.
Exam Tip: One time cannot have multiple positions.
NEET Exam-style Question 11A straight line x-t graph has positive constant slope. What does it represent?
Solution
Slope of x-t graph is velocity and constant positive slope means constant positive velocity.
Final Answer
Uniform motion in positive direction.
Exam Tip: Slope tells velocity.
NEET Exam-style Question 12A v-t graph is parallel to time axis above origin. Find acceleration.
Solution
Velocity is constant, so slope of v-t graph is zero.
Final Answer
Zero acceleration.
Exam Tip: Horizontal v-t graph means uniform velocity.
NEET Exam-style Question 13Area under a v-t graph between two times gives what physical quantity?
Solution
Small area vΔt represents displacement; total area gives displacement.
Final Answer
Displacement.
Exam Tip: Signed area matters.
NEET Exam-style Question 14Area under an a-t graph gives what physical quantity?
Solution
Since a=Δv/Δt, aΔt gives change in velocity.
Final Answer
Change in velocity.
Exam Tip: Do not confuse with displacement.
NEET Exam-style Question 15A particle returns to starting point. What is its displacement?
Solution
Initial and final positions are same, so displacement is zero even if distance is non-zero.
Final Answer
Zero.
Exam Tip: Distance and displacement differ.
NEET Exam-style Question 16For a speed-time graph, what does area represent?
Solution
Speed is non-negative, so area gives total path length.
Final Answer
Distance travelled.
Exam Tip: For speed-time graph, area is distance.
NEET Exam-style Question 17In an x-t graph, velocity is zero at a turning point. Why?
Solution
Velocity equals slope of x-t graph, and at a smooth maximum or minimum the slope is zero.
Final Answer
Tangent is horizontal.
Exam Tip: Horizontal tangent means zero instantaneous velocity.
NEET Exam-style Question 18Can acceleration be negative while speed increases?
Solution
If velocity and acceleration have the same sign, speed increases. Thus negative acceleration with negative velocity increases speed.
Final Answer
Yes, if velocity is negative.
Exam Tip: Compare signs of v and a.
JEE Main PYQs and Exam-style Questions
Authentic year is written only when certain; otherwise the card is clearly marked exam-style.
JEE Main Exam-style Question 1A straight line x-t graph has positive constant slope. What does it represent?
Solution
Slope of x-t graph is velocity and constant positive slope means constant positive velocity.
Final Answer
Uniform motion in positive direction.
Exam Tip: Slope tells velocity.
JEE Main Exam-style Question 2A v-t graph is parallel to time axis above origin. Find acceleration.
Solution
Velocity is constant, so slope of v-t graph is zero.
Final Answer
Zero acceleration.
Exam Tip: Horizontal v-t graph means uniform velocity.
JEE Main Exam-style Question 3Area under a v-t graph between two times gives what physical quantity?
Solution
Small area vΔt represents displacement; total area gives displacement.
Final Answer
Displacement.
Exam Tip: Signed area matters.
JEE Main Exam-style Question 4Area under an a-t graph gives what physical quantity?
Solution
Since a=Δv/Δt, aΔt gives change in velocity.
Final Answer
Change in velocity.
Exam Tip: Do not confuse with displacement.
JEE Main Exam-style Question 5A particle returns to starting point. What is its displacement?
Solution
Initial and final positions are same, so displacement is zero even if distance is non-zero.
Final Answer
Zero.
Exam Tip: Distance and displacement differ.
JEE Main Exam-style Question 6For a speed-time graph, what does area represent?
Solution
Speed is non-negative, so area gives total path length.
Final Answer
Distance travelled.
Exam Tip: For speed-time graph, area is distance.
JEE Main Exam-style Question 7In an x-t graph, velocity is zero at a turning point. Why?
Solution
Velocity equals slope of x-t graph, and at a smooth maximum or minimum the slope is zero.
Final Answer
Tangent is horizontal.
Exam Tip: Horizontal tangent means zero instantaneous velocity.
JEE Main Exam-style Question 8Can acceleration be negative while speed increases?
Solution
If velocity and acceleration have the same sign, speed increases. Thus negative acceleration with negative velocity increases speed.
Final Answer
Yes, if velocity is negative.
Exam Tip: Compare signs of v and a.
JEE Main Exam-style Question 9A v-t graph crosses the time axis. What happens to velocity at crossing?
Solution
The sign of velocity changes when graph crosses the time axis.
Final Answer
Velocity is zero and direction may reverse.
Exam Tip: Crossing v=0 often indicates reversal.
JEE Main Exam-style Question 10What does a vertical line on an x-t graph imply?
Solution
A vertical line gives many positions at the same time, impossible for one-dimensional motion of one particle.
Final Answer
It is impossible for one particle.
Exam Tip: One time cannot have multiple positions.
JEE Main Exam-style Question 11A straight line x-t graph has positive constant slope. What does it represent?
Solution
Slope of x-t graph is velocity and constant positive slope means constant positive velocity.
Final Answer
Uniform motion in positive direction.
Exam Tip: Slope tells velocity.
JEE Main Exam-style Question 12A v-t graph is parallel to time axis above origin. Find acceleration.
Solution
Velocity is constant, so slope of v-t graph is zero.
Final Answer
Zero acceleration.
Exam Tip: Horizontal v-t graph means uniform velocity.
JEE Main Exam-style Question 13Area under a v-t graph between two times gives what physical quantity?
Solution
Small area vΔt represents displacement; total area gives displacement.
Final Answer
Displacement.
Exam Tip: Signed area matters.
JEE Main Exam-style Question 14Area under an a-t graph gives what physical quantity?
Solution
Since a=Δv/Δt, aΔt gives change in velocity.
Final Answer
Change in velocity.
Exam Tip: Do not confuse with displacement.
JEE Main Exam-style Question 15A particle returns to starting point. What is its displacement?
Solution
Initial and final positions are same, so displacement is zero even if distance is non-zero.
Final Answer
Zero.
Exam Tip: Distance and displacement differ.
JEE Main Exam-style Question 16For a speed-time graph, what does area represent?
Solution
Speed is non-negative, so area gives total path length.
Final Answer
Distance travelled.
Exam Tip: For speed-time graph, area is distance.
JEE Main Exam-style Question 17In an x-t graph, velocity is zero at a turning point. Why?
Solution
Velocity equals slope of x-t graph, and at a smooth maximum or minimum the slope is zero.
Final Answer
Tangent is horizontal.
Exam Tip: Horizontal tangent means zero instantaneous velocity.
JEE Main Exam-style Question 18Can acceleration be negative while speed increases?
Solution
If velocity and acceleration have the same sign, speed increases. Thus negative acceleration with negative velocity increases speed.
Final Answer
Yes, if velocity is negative.
Exam Tip: Compare signs of v and a.
JEE Advanced PYQs and Exam-style Questions
Authentic year is written only when certain; otherwise the card is clearly marked exam-style.
JEE Advanced Exam-style Question 1A straight line x-t graph has positive constant slope. What does it represent?
Solution
Slope of x-t graph is velocity and constant positive slope means constant positive velocity.
Final Answer
Uniform motion in positive direction.
Exam Tip: Slope tells velocity.
JEE Advanced Exam-style Question 2A v-t graph is parallel to time axis above origin. Find acceleration.
Solution
Velocity is constant, so slope of v-t graph is zero.
Final Answer
Zero acceleration.
Exam Tip: Horizontal v-t graph means uniform velocity.
JEE Advanced Exam-style Question 3Area under a v-t graph between two times gives what physical quantity?
Solution
Small area vΔt represents displacement; total area gives displacement.
Final Answer
Displacement.
Exam Tip: Signed area matters.
JEE Advanced Exam-style Question 4Area under an a-t graph gives what physical quantity?
Solution
Since a=Δv/Δt, aΔt gives change in velocity.
Final Answer
Change in velocity.
Exam Tip: Do not confuse with displacement.
JEE Advanced Exam-style Question 5A particle returns to starting point. What is its displacement?
Solution
Initial and final positions are same, so displacement is zero even if distance is non-zero.
Final Answer
Zero.
Exam Tip: Distance and displacement differ.
JEE Advanced Exam-style Question 6For a speed-time graph, what does area represent?
Solution
Speed is non-negative, so area gives total path length.
Final Answer
Distance travelled.
Exam Tip: For speed-time graph, area is distance.
JEE Advanced Exam-style Question 7In an x-t graph, velocity is zero at a turning point. Why?
Solution
Velocity equals slope of x-t graph, and at a smooth maximum or minimum the slope is zero.
Final Answer
Tangent is horizontal.
Exam Tip: Horizontal tangent means zero instantaneous velocity.
JEE Advanced Exam-style Question 8Can acceleration be negative while speed increases?
Solution
If velocity and acceleration have the same sign, speed increases. Thus negative acceleration with negative velocity increases speed.
Final Answer
Yes, if velocity is negative.
Exam Tip: Compare signs of v and a.
JEE Advanced Exam-style Question 9A v-t graph crosses the time axis. What happens to velocity at crossing?
Solution
The sign of velocity changes when graph crosses the time axis.
Final Answer
Velocity is zero and direction may reverse.
Exam Tip: Crossing v=0 often indicates reversal.
JEE Advanced Exam-style Question 10What does a vertical line on an x-t graph imply?
Solution
A vertical line gives many positions at the same time, impossible for one-dimensional motion of one particle.
Final Answer
It is impossible for one particle.
Exam Tip: One time cannot have multiple positions.
JEE Advanced Exam-style Question 11A straight line x-t graph has positive constant slope. What does it represent?
Solution
Slope of x-t graph is velocity and constant positive slope means constant positive velocity.
Final Answer
Uniform motion in positive direction.
Exam Tip: Slope tells velocity.
JEE Advanced Exam-style Question 12A v-t graph is parallel to time axis above origin. Find acceleration.
Solution
Velocity is constant, so slope of v-t graph is zero.
Final Answer
Zero acceleration.
Exam Tip: Horizontal v-t graph means uniform velocity.
Assertion Reason: 30 Questions
Assertion Reason 1Assertion: The slope of x-t graph gives velocity. Reason: Velocity is rate of change of position with time.
Solution
Both assertion and reason are true, and reason explains assertion.
Final Answer
Both assertion and reason are true, and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 2Assertion: Area under v-t graph gives acceleration. Reason: Velocity multiplied by time has unit of displacement.
Solution
Assertion is false; reason is true. Area under v-t graph gives displacement.
Final Answer
Assertion is false; reason is true. Area under v-t graph gives displacement.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 3Assertion: Average speed is always greater than or equal to magnitude of average velocity. Reason: Total path length is always greater than or equal to displacement magnitude.
Solution
Both are true and reason explains assertion.
Final Answer
Both are true and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 4Assertion: A horizontal v-t graph means zero acceleration. Reason: Acceleration is slope of v-t graph.
Solution
Both are true and reason explains assertion.
Final Answer
Both are true and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 5Assertion: Speed can be negative. Reason: Velocity can be negative.
Solution
Assertion is false; reason is true. Speed is magnitude of velocity.
Final Answer
Assertion is false; reason is true. Speed is magnitude of velocity.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 6Assertion: The slope of x-t graph gives velocity. Reason: Velocity is rate of change of position with time.
Solution
Both assertion and reason are true, and reason explains assertion.
Final Answer
Both assertion and reason are true, and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 7Assertion: Area under v-t graph gives acceleration. Reason: Velocity multiplied by time has unit of displacement.
Solution
Assertion is false; reason is true. Area under v-t graph gives displacement.
Final Answer
Assertion is false; reason is true. Area under v-t graph gives displacement.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 8Assertion: Average speed is always greater than or equal to magnitude of average velocity. Reason: Total path length is always greater than or equal to displacement magnitude.
Solution
Both are true and reason explains assertion.
Final Answer
Both are true and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 9Assertion: A horizontal v-t graph means zero acceleration. Reason: Acceleration is slope of v-t graph.
Solution
Both are true and reason explains assertion.
Final Answer
Both are true and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 10Assertion: Speed can be negative. Reason: Velocity can be negative.
Solution
Assertion is false; reason is true. Speed is magnitude of velocity.
Final Answer
Assertion is false; reason is true. Speed is magnitude of velocity.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 11Assertion: The slope of x-t graph gives velocity. Reason: Velocity is rate of change of position with time.
Solution
Both assertion and reason are true, and reason explains assertion.
Final Answer
Both assertion and reason are true, and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 12Assertion: Area under v-t graph gives acceleration. Reason: Velocity multiplied by time has unit of displacement.
Solution
Assertion is false; reason is true. Area under v-t graph gives displacement.
Final Answer
Assertion is false; reason is true. Area under v-t graph gives displacement.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 13Assertion: Average speed is always greater than or equal to magnitude of average velocity. Reason: Total path length is always greater than or equal to displacement magnitude.
Solution
Both are true and reason explains assertion.
Final Answer
Both are true and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 14Assertion: A horizontal v-t graph means zero acceleration. Reason: Acceleration is slope of v-t graph.
Solution
Both are true and reason explains assertion.
Final Answer
Both are true and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 15Assertion: Speed can be negative. Reason: Velocity can be negative.
Solution
Assertion is false; reason is true. Speed is magnitude of velocity.
Final Answer
Assertion is false; reason is true. Speed is magnitude of velocity.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 16Assertion: The slope of x-t graph gives velocity. Reason: Velocity is rate of change of position with time.
Solution
Both assertion and reason are true, and reason explains assertion.
Final Answer
Both assertion and reason are true, and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 17Assertion: Area under v-t graph gives acceleration. Reason: Velocity multiplied by time has unit of displacement.
Solution
Assertion is false; reason is true. Area under v-t graph gives displacement.
Final Answer
Assertion is false; reason is true. Area under v-t graph gives displacement.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 18Assertion: Average speed is always greater than or equal to magnitude of average velocity. Reason: Total path length is always greater than or equal to displacement magnitude.
Solution
Both are true and reason explains assertion.
Final Answer
Both are true and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 19Assertion: A horizontal v-t graph means zero acceleration. Reason: Acceleration is slope of v-t graph.
Solution
Both are true and reason explains assertion.
Final Answer
Both are true and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 20Assertion: Speed can be negative. Reason: Velocity can be negative.
Solution
Assertion is false; reason is true. Speed is magnitude of velocity.
Final Answer
Assertion is false; reason is true. Speed is magnitude of velocity.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 21Assertion: The slope of x-t graph gives velocity. Reason: Velocity is rate of change of position with time.
Solution
Both assertion and reason are true, and reason explains assertion.
Final Answer
Both assertion and reason are true, and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 22Assertion: Area under v-t graph gives acceleration. Reason: Velocity multiplied by time has unit of displacement.
Solution
Assertion is false; reason is true. Area under v-t graph gives displacement.
Final Answer
Assertion is false; reason is true. Area under v-t graph gives displacement.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 23Assertion: Average speed is always greater than or equal to magnitude of average velocity. Reason: Total path length is always greater than or equal to displacement magnitude.
Solution
Both are true and reason explains assertion.
Final Answer
Both are true and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 24Assertion: A horizontal v-t graph means zero acceleration. Reason: Acceleration is slope of v-t graph.
Solution
Both are true and reason explains assertion.
Final Answer
Both are true and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 25Assertion: Speed can be negative. Reason: Velocity can be negative.
Solution
Assertion is false; reason is true. Speed is magnitude of velocity.
Final Answer
Assertion is false; reason is true. Speed is magnitude of velocity.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 26Assertion: The slope of x-t graph gives velocity. Reason: Velocity is rate of change of position with time.
Solution
Both assertion and reason are true, and reason explains assertion.
Final Answer
Both assertion and reason are true, and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 27Assertion: Area under v-t graph gives acceleration. Reason: Velocity multiplied by time has unit of displacement.
Solution
Assertion is false; reason is true. Area under v-t graph gives displacement.
Final Answer
Assertion is false; reason is true. Area under v-t graph gives displacement.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 28Assertion: Average speed is always greater than or equal to magnitude of average velocity. Reason: Total path length is always greater than or equal to displacement magnitude.
Solution
Both are true and reason explains assertion.
Final Answer
Both are true and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 29Assertion: A horizontal v-t graph means zero acceleration. Reason: Acceleration is slope of v-t graph.
Solution
Both are true and reason explains assertion.
Final Answer
Both are true and reason explains assertion.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Assertion Reason 30Assertion: Speed can be negative. Reason: Velocity can be negative.
Solution
Assertion is false; reason is true. Speed is magnitude of velocity.
Final Answer
Assertion is false; reason is true. Speed is magnitude of velocity.
Exam Tip: Use standard AR logic: check assertion first, then reason, then explanation.
Case Studies Based on Motion Graphs
Case Study 1Position-time graph of two students
Passage
Two students return home from school. Student A starts earlier but has a smaller slope on the x-t graph. Student B starts later but has a larger slope and the two lines intersect once.
Questions
Who is faster? Who starts earlier? What does the intersection mean?
Answers
B is faster; A starts earlier; intersection means B overtakes A.
Explanation
In x-t graph, slope gives velocity and intersection means same position at same time.
Exam Tip: Always identify the graph type before interpreting slope or area.
Case Study 2Velocity-time graph of a train
Passage
A train starts from rest, accelerates uniformly, moves with constant velocity, and then retards uniformly to rest.
Questions
Where is acceleration positive, zero and negative? What does area give?
Answers
Positive in first part, zero in middle, negative in last part. Area gives displacement.
Explanation
Slope of v-t gives acceleration; area gives displacement.
Exam Tip: Always identify the graph type before interpreting slope or area.
Case Study 3Acceleration-time graph pulse
Passage
A ball is hit by a bat. Its acceleration-time graph has a sharp narrow positive pulse.
Questions
What does area under the pulse represent?
Answers
Change in velocity due to impulse-like contact.
Explanation
Short large acceleration can produce significant velocity change.
Exam Tip: Always identify the graph type before interpreting slope or area.
Case Study 4Free-fall graph
Passage
A stone is dropped from rest. Take downward positive.
Questions
Sketch v-t and a-t meanings.
Answers
v-t is a straight line with positive slope g; a-t is horizontal at +g.
Explanation
Sign convention decides graph sign.
Exam Tip: Always identify the graph type before interpreting slope or area.
Case Study 5Bouncing ball speed graph
Passage
A ball bounces and loses speed after each collision.
Questions
Why does speed-time graph show sudden drops?
Answers
Collision is brief and speed immediately after bounce is smaller.
Explanation
Speed is never negative, but it can jump discontinuously in an ideal collision.
Exam Tip: Always identify the graph type before interpreting slope or area.
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Motion in a Straight Line
A complete premium roadmap for position, speed, velocity, acceleration, equations of motion, free fall and motion graphs for CBSE, NEET, JEE, IB, IGCSE and A-Level Physics.
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Chapter Roadmap
Each card opens the complete notes page. Click anywhere on the card: title, illustration, chips, description or background.
Position, Distance and Displacement
Start with origin, coordinate axis, position, path length and signed displacement.
OriginPositionDistanceDisplacement
Open Notes
Speed and Velocity
Compare scalar speed with vector velocity using average and instantaneous quantities.
SpeedVelocityAverageInstantaneous
Open Notes
Acceleration and Retardation
Understand change in velocity, positive acceleration, negative acceleration and motion direction.
AccelerationRetardationSignsUnits
Open Notes
Equations of Motion and Free Fall
Derive v = u + at, s = ut + ½at2, v2 = u2 + 2as and apply gravity signs.
DerivationsFree FallGalileoNumericals
Open Notes
Graphs of Motion
Read x-t, v-t and a-t graphs using slope, area, intercepts and sign of motion.
x-t Graphv-t Grapha-t GraphArea
Open Notes
Why Study Motion in a Straight Line?
Direct formula-based questions, free fall, graphs and units appear frequently in medical entrance preparation.
One-dimensional kinematics builds speed for mixed numerical problems in mechanics.
Advanced questions combine graphs, sign convention, relative motion and calculus-based reasoning.
Newton's laws, friction, circular motion and projectile motion require strong kinematics basics.
Displacement, velocity and acceleration connect directly with work, kinetic energy and power.
Linear motion concepts become angular displacement, angular velocity and angular acceleration.