Class 11 · Physics · Chapter 2
Motion in a Straight Line
This chapter builds the language of one-dimensional motion — position, displacement, velocity, acceleration and the three kinematic equations — and teaches you to read position-time and velocity-time graphs. There is no Class 11 board exam, but this chapter is tested in the school final and it is the base every kinematics question in Class 12, JEE and NEET is built on.
Before you startWhat the board asks from this chapter
No CBSE board exam in Class 11. In the school final (and in most schools' unit tests), this chapter usually carries about 5–7 marks out of the mechanics unit, often as one graph-based question plus one numerical on the equations of motion.
| Question type | Marks | Usually asked |
|---|---|---|
| MCQ | 1 | Sign of velocity/acceleration from a graph, whether displacement can be zero while distance is not, or a quick numerical using v = u + at. |
| Assertion–Reason | 1 | Tests whether you understand *why* a rule holds — e.g. why speed is never negative but velocity can be, or why acceleration can be nonzero at zero velocity (as at the top of a vertical throw). |
| Very short answer | 2 | Define a term, or read one value (velocity/acceleration) off a graph at a given instant. |
| Short answer | 3 | A numerical using the three kinematic equations, or a relative-velocity problem with two bodies moving on the same line. |
| Long answer | 5 | Derive the three kinematic equations by calculus or graphically, then apply one of them to a free-fall or braking-distance scenario. |
| Case-based | 4 | A short passage (a car braking, a ball thrown up, two trains) followed by 4 sub-questions mixing concept and calculation. |
Hot topics, in order
- Distance vs displacement, and when displacement is zero but distance is not
- Average velocity vs average speed vs instantaneous velocity
- Reading slope on a position–time graph and area under a velocity–time graph
- Deriving v = u + at, s = ut + ½at², v² = u² + 2as (both calculus and graphical methods)
- Motion under gravity: time to reach maximum height, maximum height, total time of flight
- Relative velocity of two objects moving on the same straight line
- Stopping distance and reaction time numericals
- Interpreting whether a v–t or a–t graph is physically possible
The whole chapter, in orderNotes
01 2.1 Introduction: Motion, Point Object and Frame of Reference
A body is said to be in motion if its position changes with time relative to a chosen reference point, called the origin, fixed in a frame of reference. Motion is always described relative to some observer — a passenger sitting still in a moving train is at rest with respect to the train but in motion with respect to the ground. There is no such thing as absolute motion.
When the size of a body is much smaller than the distance it travels, we treat it as a point object — its internal structure and rotation don't matter, only the position of one point. A train travelling several hundred kilometres can be treated as a point object for that journey, even though it cannot be treated as a point object while it is crossing a platform.
Motion in a straight line — also called rectilinear motion or one-dimensional motion — is the simplest case, where the object's position can be described by a single coordinate x measured along the line, with a sign showing direction. This chapter fixes the vocabulary (position, displacement, velocity, acceleration) that every later mechanics chapter reuses.
02 2.2 Position, Path Length and Displacement
Position is the location of the object with respect to the chosen origin, given a sign convention (usually right/up = positive). Path length (or distance) is the total length of the actual path travelled — it is a scalar and can only increase or stay the same. Displacement is the change in position, Δx = x2 − x1; it is a vector, so in one dimension it carries a sign showing direction, and it is the straight-line (shortest) separation between the initial and final position.
Displacement can be zero even when the path length is not: a body that returns exactly to its starting point has covered a nonzero distance but zero displacement. This is the single most repeated conceptual MCQ from this chapter. Displacement magnitude can never exceed the path length; the two are equal only when the motion is along a straight line without any reversal in direction.
Example: a man walks 8 m east then 6 m west. Path length = 8 + 6 = 14 m. Displacement = 8 − 6 = 2 m east (taking east as positive).
03 2.3 Average Velocity and Average Speed
Average velocity is displacement divided by the time interval over which it occurs; it is a vector and can be positive, negative or zero. Average speed is total path length divided by total time; it is a scalar and is always positive (or zero only if the body never moved).
Average speed is greater than or equal to the magnitude of average velocity, with equality only when the motion is one-directional (no reversal, no idle stop that isn't counted in the path). For a round trip, average velocity is zero but average speed is not, because the denominator (time) is the same but the numerator changes from displacement to path length.
Worked example: a car covers 100 km in 2 hours going out and 100 km in 3 hours coming back along the same road. Average speed = total distance / total time = 200/5 = 40 km/h. Average velocity = total displacement / total time = 0/5 = 0, since the car ends where it started.
04 2.4 Instantaneous Velocity and Speed
Instantaneous velocity is the velocity of an object at one particular instant of time — the limit of average velocity as the time interval Δt shrinks to zero. Mathematically it is the derivative of position with respect to time: v = dx/dt. Geometrically, on a position–time graph, the instantaneous velocity at any point is the slope of the tangent to the curve at that point.
Instantaneous speed is simply the magnitude of the instantaneous velocity, |v|. Unlike average speed and average velocity, instantaneous speed and instantaneous velocity always have the same magnitude at every instant — there is no averaging over a path to create a difference between them.
Reading a position–time (x–t) graph is a core exam skill: a straight-line x–t graph means constant velocity (uniform motion); a curved x–t graph means the velocity is changing (non-uniform, accelerated motion); a horizontal line means the object is at rest; and a steeper slope always means a larger speed, regardless of the sign of the slope.
05 2.5 Acceleration
Average acceleration is the change in velocity divided by the time taken: aavg = Δv/Δt. Instantaneous acceleration is the limit of this ratio as Δt → 0, i.e. a = dv/dt, the slope of the tangent to the velocity–time (v–t) graph at that instant. Since v = dx/dt, acceleration is also the second derivative of position: a = d²x/dt².
Acceleration is a vector; its sign shows direction, not simply 'speeding up' or 'slowing down'. An object speeds up when velocity and acceleration have the same sign, and slows down (decelerates) when they have opposite signs — regardless of whether acceleration itself is called 'positive' or 'negative' in the chosen convention. A body can have nonzero acceleration even at the instant its velocity is zero: at the highest point of a ball thrown straight up, v = 0 but a = −g (downward) throughout, because gravity never switches off.
On a v–t graph: the slope gives acceleration, and the area under the curve (between the curve and the time axis) gives the displacement over that interval. This area rule works even when acceleration is changing, because it comes directly from a = dv/dt and v = dx/dt.
06 2.6 Position–Time and Velocity–Time Graphs
Graph questions are asked almost every year because they test understanding, not memorised formulas. Keep these rules fixed in mind:
- On an x–t graph: slope = velocity. A straight line = uniform velocity; increasing slope with time = increasing velocity (accelerated motion); a horizontal segment = object at rest; a line crossing the time-axis means the object passes through the origin, not that it stops.
- On a v–t graph: slope = acceleration; area between the curve and the time-axis = displacement. A straight horizontal line = zero acceleration (uniform velocity); a straight line with constant nonzero slope = uniform acceleration; area below the time axis counts as negative displacement.
- On an a–t graph: area under the curve = change in velocity, Δv.
A v–t graph cannot be discontinuous with a sudden vertical jump in real motion (that would need infinite acceleration), and it cannot show two different velocities at the same instant of time — both are common 'which graph is not possible' MCQ traps. Similarly an x–t graph can never fold back on itself (be double-valued in x for one value of t), since an object cannot be at two positions at once.
07 2.7 Kinematic Equations for Uniformly Accelerated Motion
When acceleration a is constant, three equations connect initial velocity u, final velocity v, time t, acceleration a and displacement s. These can be derived either graphically (from the area and slope of a straight-line v–t graph) or by calculus (integrating a = dv/dt and v = dx/dt).
Calculus derivation of v = u + at: a = dv/dt is constant, so ∫uv dv = ∫0t a dt, giving v − u = at, i.e. v = u + at.
Calculus derivation of s = ut + ½at²: v = dx/dt = u + at, so ∫0s dx = ∫0t (u + at) dt, giving s = ut + ½at².
Calculus derivation of v² = u² + 2as: a = v(dv/dx) (using the chain rule, since a = dv/dt = (dv/dx)(dx/dt) = v dv/dx). So a dx = v dv, and integrating both sides, a∫0s dx = ∫uv v dv gives as = (v² − u²)/2, i.e. v² = u² + 2as.
A fourth useful (non-NCERT-numbered but frequently used) relation gives the distance covered in the nth second: sn = u + a(n − ½), obtained by subtracting the displacement up to (n−1) seconds from the displacement up to n seconds. These equations hold only when acceleration is uniform (constant in both magnitude and direction); for non-uniform acceleration you must go back to calculus or graphs.
08 2.8 Motion Under Gravity: Free Fall as a Special Case
Free fall (dropping) and vertical throw (projected upward) are the most common application of the kinematic equations in Class 11 exams, because gravity provides a constant, known acceleration g ≈ 9.8 m s−2 directed downward (many textbooks and exam solutions round this to 10 m s−2 for simpler arithmetic; state the value you use).
Taking upward as positive: for a ball thrown up with initial speed u, acceleration is a = −g throughout the flight (even at the highest point, where v = 0 momentarily but a = −g never becomes zero). Time to reach maximum height, tup = u/g. Maximum height, hmax = u²/2g. If the ball returns to the point of projection, total time of flight T = 2u/g, and by symmetry the time to go up equals the time to come down, and the speed of return equals the launch speed u (ignoring air resistance).
Worked example: a ball is thrown vertically upward with u = 20 m s−1 (take g = 10 m s−2). Time to reach the top = u/g = 20/10 = 2 s. Maximum height = u²/2g = 400/20 = 20 m. Total time of flight = 2u/g = 4 s. Checking with s = ut + ½at² at t = 2 s: s = 20(2) − ½(10)(4) = 40 − 20 = 20 m — matches.
09 2.9 Relative Velocity
The relative velocity of object A with respect to object B is the velocity A appears to have as seen by an observer moving with B: vAB = vA − vB. In one dimension this is a straightforward subtraction with sign, but the sign convention must be applied carefully.
If two objects move in the same direction along a line with speeds vA and vB (vA > vB), the relative velocity of A with respect to B is vA − vB, and this is also the rate at which A gains on B — useful for 'time to overtake' problems. If they move in opposite directions, taking one direction as positive, the relative velocity is vA − (−vB) = vA + vB, i.e. the speeds add — this is why two vehicles approaching each other close the gap much faster than either one's own speed.
Worked example: two trains, each 100 m long, move on parallel tracks at 54 km/h and 36 km/h. (a) In the same direction: relative velocity = 54 − 36 = 18 km/h = 5 m/s; time to cross each other = (100+100)/5 = 40 s. (b) In opposite directions: relative velocity = 54 + 36 = 90 km/h = 25 m/s; time to cross = 200/25 = 8 s.
VocabularyKey terms
Keep this on one pageFormula sheet
| Name | Expression | Notes |
|---|---|---|
| Displacement | Δx = x2 − x1 | vector; can be zero even when distance travelled is not |
| Average velocity | vavg = Δx/Δt | displacement over total time; can be negative or zero |
| Average speed | speedavg = total path length/Δt | always ≥ |vavg|; use total distance/total time for multi-leg journeys, never average the speeds directly |
| Instantaneous velocity | v = dx/dt | slope of tangent to x–t graph |
| Instantaneous acceleration | a = dv/dt = d2x/dt2 | slope of tangent to v–t graph |
| First equation of motion | v = u + at | valid only for constant acceleration |
| Second equation of motion | s = ut + ½at2 | displacement in time t under constant acceleration |
| Third equation of motion | v2 = u2 + 2as | connects velocity and displacement, no time term |
| Distance in nth second | sn = u + a(n − ½) | n is an integer count of the second, not a continuous time |
| Free fall — max height | hmax = u2/2g | for a body thrown up with speed u; g ≈ 9.8 m s−2 (often taken as 10) |
| Free fall — time of flight | T = 2u/g | time to return to the point of projection, ignoring air resistance |
| Relative velocity (same line) | vAB = vA − vB | subtract for same direction, add magnitudes for opposite directions |
Practise with a pencilDiagrams that earn marks
Position–time (x–t) graphs for different types of motion
Draw four small x–t graphs side by side on the same x and t axes: (1) a horizontal straight line labelled 'at rest', (2) a straight line with positive slope labelled 'uniform velocity', (3) an upward-curving line (slope increasing with time) labelled 'uniformly accelerated, starting from rest', (4) a straight line with negative slope labelled 'uniform velocity in the negative direction'. Mark the slope at one point on graph 3 with a tangent line and label it 'instantaneous velocity = slope of tangent'.
Velocity–time (v–t) graph for uniformly accelerated motion, with the area used to derive s = ut + ½at²
Draw a straight line starting at (0, u) on the v-axis and rising to (t, v) at time t, where v = u + at. Shade the trapezium area under the line between t = 0 and t. Split the trapezium into a rectangle of height u and width t (area = ut) sitting below a triangle of base t and height (v − u) = at (area = ½at²). Label the total shaded area as displacement s = ut + ½at², matching the algebraic derivation.
Ball thrown vertically upward — velocity and acceleration at three points
Draw a vertical line showing the path of the ball with three marked points: the launch point (velocity u, upward arrow), the highest point (velocity = 0, but still draw a downward arrow labelled 'a = −g, never zero'), and the point of return (velocity u, downward arrow, same magnitude as launch by symmetry). Add a small v–t graph alongside it: a straight line starting at +u, crossing zero at t = u/g, and reaching −u at t = 2u/g, with constant negative slope −g throughout.
TextbookNCERT questions the board keeps asking
2.1 In 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?2 m
(a) Yes — the carriage travels a distance (the length of the track) much larger than its own size, so it can be treated as a point object.
(b) Yes — the monkey and cyclist together travel a distance around the track much larger than their size.
(c) No — the ball's own spin and the sharp turn it takes on hitting the ground depend on its size and orientation, which cannot be ignored.
(d) No — the beaker tumbles over a distance comparable to its own size, so its shape and orientation matter and it cannot be treated as a point.
2.2 The 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. 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).3 m
(a) A lives closer to the school than B (A's line reaches a smaller final x value / P is nearer O than Q).
(b) A starts from the school earlier than B.
(c) B walks faster than A (B's x–t line has a steeper slope).
(d) A and B reach home at different times.
(e) B overtakes A on the road once (B starts later but, being faster, crosses A's path once before reaching home).
2.3 A woman starts from her home at 9:00 am, walks with a speed of 5 km/h on a straight road up to her office 2.5 km away, stays at the office until 5:00 pm, and returns home by an auto with a speed of 25 km/h. Choose suitable scales and plot the x–t graph of her motion.3 m
Time to walk to office = distance/speed = 2.5/5 = 0.5 h = 30 min, so she reaches office at 9:30 am. The graph rises from x = 0 at t = 9:00 am to x = 2.5 km at t = 9:30 am, a straight line of positive slope 5 km/h. It then stays flat at x = 2.5 km from 9:30 am to 5:00 pm (she is stationary at office). Time for the return auto ride = 2.5/25 = 0.1 h = 6 min, so she is home by 5:06 pm; the graph falls steeply back to x = 0 between 5:00 pm and 5:06 pm, a straight line of slope −25 km/h (much steeper than the outward walk since the auto is faster).
2.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, and so on, each step 1 m long and taking 1 s. Effectively he moves 8 m in 9 s (why?). Determine how long the drunkard takes to fall in a pit 13 m away from the start.3 m
In every block of 8 s (5 forward + 3 backward steps), his net displacement is 5 − 3 = 2 m, so effectively he moves 2 m every 8 s.
After 4 such blocks (32 s), his net displacement = 4 × 2 = 8 m. In the next cycle he takes 5 more forward steps (5 s), reaching 8 + 5 = 13 m at the pit — so he falls into the pit exactly when his forward steps carry him there, without needing to step back. Time taken = 32 + 5 = 37 s.
2.5 A car moving along a straight highway with a speed of 126 km/h 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?3 m
u = 126 km/h = 126 × (5/18) = 35 m/s. v = 0, s = 200 m.
Using v2 = u2 + 2as: 0 = (35)2 + 2a(200) ⟹ a = −1225/400 = −3.0625 m/s2, so retardation ≈ 3.06 m/s2.
Using v = u + at: 0 = 35 + (−3.0625)t ⟹ t = 35/3.0625 ≈ 11.4 s.
2.6 A player throws a ball upwards with an initial speed of 29.4 m/s. (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 x-axis, and give the signs of position, velocity and acceleration of the ball during its 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/s2 and neglect air resistance.)5 m
(a) The acceleration is directed vertically downward (towards the ground) throughout the flight, since it is due to gravity alone; it does not reverse direction just because the ball is going up.
(b) At the highest point, velocity = 0, but acceleration = 9.8 m/s2 downward (unchanged) — velocity being zero does not mean acceleration is zero.
(c) With downward taken as positive and the highest point as origin: as the ball falls, position x is positive and increasing, velocity is positive and increasing, and acceleration is positive (= +g) throughout.
(d) Maximum height h = u2/2g = (29.4)2/(2 × 9.8) = 864.36/19.6 = 44.1 m. Total time of flight T = 2u/g = 2(29.4)/9.8 = 6.0 s.
2.7 Explain 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. 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.)3 m
(a) Magnitude of displacement is the straight-line distance between the start and end position; path length is the total distance actually walked. Example: a person walks 5 m east then 3 m west — path length = 8 m, but magnitude of displacement = 2 m. Path length ≥ |displacement| always, because any deviation or reversal in the path only adds to the length walked without adding proportionally to the net separation.
(b) Average speed = path length/time, average velocity magnitude = |displacement|/time. Since path length ≥ |displacement| for the same time interval, average speed ≥ |average velocity|.
Equality holds in both cases only when the motion is strictly one-directional (the particle never reverses direction), so that path length exactly equals the magnitude of displacement.
2.8 A car moves from X to Y with a uniform speed vu and returns to X with a uniform speed vd. What is the average speed for this round trip?3 m
Let the one-way distance be d. Time for X→Y = d/vu; time for Y→X = d/vd. Total distance = 2d; total time = d/vu + d/vd = d(vu+vd)/(vuvd).
Average speed = total distance/total time = 2d ÷ [d(vu+vd)/(vuvd)] = 2vuvd/(vu+vd). This is the harmonic mean of the two speeds, not their simple average — a common exam trap.
2.9 Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km/h in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?5 m
Both trains start with the same speed 72 km/h = 20 m/s, so B's initial velocity relative to A is 0. Since A moves at constant velocity, B's acceleration relative to A equals B's own acceleration, a = 1 m/s2.
Relative displacement of B's front with respect to A's front in t = 50 s: s = 0(50) + ½(1)(50)2 = 1250 m.
This 1250 m is the distance B's front gains on A's front. For B's guard (at the rear of B, 400 m behind B's front) to just reach A's driver (at the front of A), B's front must have advanced 1250 m relative to A's front, which equals: (original gap between B's front and A's rear) + (length of A, 400 m, since B's front must pass all of A) − (length of B, 400 m, since it is the rear of B, not the front, that needs to reach A's front). These two 400 m lengths cancel, so the original gap between the two trains = 1250 m − 400 m = 850 m (subtracting train A's length, following the standard NCERT solution method).
2.10 On a two-lane road, car A is travelling with a speed of 36 km/h. Two cars B and C approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?5 m
Convert to m/s: vA = 36 km/h = 10 m/s, vB = vC = 54 km/h = 15 m/s.
Relative velocity of B with respect to A (both initially moving towards each other, B behind A in the same lane direction as A but overtaking, C approaching A from the front) = 15 − 10 = 5 m/s... Following the standard NCERT solution: relative velocity of C with respect to A (approaching, opposite direction) = 15 + 10 = 25 m/s. Time for C to cover the 1 km gap to A: t = 1000/25 = 40 s. This is the time B has to complete the overtake.
Relative velocity of B with respect to A (B behind A, same direction) = 15 − 10 = 5 m/s. In 40 s at this relative velocity, B would only close 5 × 40 = 200 m of the 1000 m gap, so B needs additional relative acceleration a to cover the full 1000 m: 1000 = 5(40) + ½a(40)2 ⟹ 1000 − 200 = 800a ⟹ a = 800/800 = 1 m/s2.
Previous years and frequently askedBoard questions with model answers
MCQ A body is thrown vertically upward. Which of the following remains constant during its flight (ignoring air resistance)?- Its velocity
- Its speed
- Its acceleration
- Its displacement
Frequently asked1 m
MCQ The area under a velocity–time graph, between the curve and the time axis, represents:- Acceleration
- Displacement
- Average speed
- Jerk
Frequently asked1 m
MCQ A particle moves in a straight line and returns to its starting point after 20 s, having covered a total path length of 60 m. Its average velocity for this interval is:- 3 m/s
- 60 m/s
- Zero
- Cannot be determined
Frequently asked1 m
MCQ Two cars A and B move in the same direction on a straight road at 20 m/s and 15 m/s respectively. The velocity of A relative to B is:- 35 m/s
- 5 m/s
- 20 m/s
- 15 m/s
Frequently asked1 m
MCQ On a position–time graph, a straight horizontal line parallel to the time axis represents:- Uniform velocity
- Uniform acceleration
- The object at rest
- Non-uniform motion
Frequently asked1 m
MCQ A ball is dropped from a height and falls freely under gravity. Its speed–time graph is:- A horizontal straight line
- A straight line through the origin with positive slope
- A parabola opening upward
- A straight line not passing through the origin, with zero slope
Frequently asked1 m
MCQ The displacement of a particle is given by x = 3t2 + 2t + 5 (SI units). Its acceleration is:- 3 m/s2
- 6 m/s2
- 2 m/s2
- 5 m/s2
Frequently asked1 m
Assertion–Reason Assertion (A): The speed of a particle can never be negative. Reason (R): Speed is the magnitude of velocity, and the magnitude of a vector quantity is always non-negative.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): A particle can have zero velocity but nonzero acceleration at the same instant. Reason (R): At the highest point of a vertically thrown ball, velocity is zero but the acceleration due to gravity continues to act.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): Average speed of a body can never be less than the magnitude of its average velocity. Reason (R): Path length covered by a body is always greater than or equal to the magnitude of its displacement.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Very short answer Can a body have zero velocity and still be accelerating? Give one example.Frequently asked2 m
Very short answer A particle moves along the x-axis such that its position is x = 2t3 − 3t2 + 4 (SI units). Find its velocity and acceleration at t = 2 s.Frequently asked2 m
a = dv/dt = 12t − 6. At t = 2 s: a = 12(2) − 6 = 24 − 6 = 18 m/s2.
Very short answer State two differences between distance and displacement.Frequently asked2 m
(2) Distance can never decrease and is always positive for a moving body; displacement can be positive, negative or zero, and its magnitude can be less than, but never more than, the distance travelled.
Very short answer What does the slope of a velocity–time graph represent? What does the area under it represent?Frequently asked2 m
Very short answer A car travels the first half of a distance between two places at a speed of 30 km/h and the second half at a speed of 50 km/h. Find the average speed of the car for the whole journey.Frequently asked2 m
Short answer Derive the equation s = ut + ½at2 for uniformly accelerated motion using the velocity–time graph.Frequently asked3 m
Split the trapezium into a rectangle of height u and base t (area = ut) and a triangle sitting above it with base t and height (v − u) (area = ½ t(v − u)).
Since a = (v − u)/t, we have (v − u) = at, so the triangle's area = ½at2.
Total area (= displacement) s = ut + ½at2.
Short answer A car starts from rest and accelerates uniformly at 2 m/s2 for 10 s. It then continues at the constant velocity reached for 20 s, and finally decelerates uniformly to rest in 5 s. Find (a) the maximum velocity reached, (b) the total distance travelled.Frequently asked3 m
(b) Stage 1 (0–10 s, accelerating): s1 = ut + ½at2 = 0 + ½(2)(100) = 100 m.
Stage 2 (10–30 s, constant 20 m/s): s2 = 20 × 20 = 400 m.
Stage 3 (last 5 s, decelerating from 20 m/s to 0): s3 = (u+v)/2 × t = (20+0)/2 × 5 = 50 m.
Total distance = 100 + 400 + 50 = 550 m.
Short answer A stone is dropped from the top of a tower 45 m high, and at the same time another stone is thrown upward from the ground with a velocity of 20 m/s along the same vertical line. Find when and where the two stones meet. (Take g = 10 m/s2.)Frequently asked3 m
Dropped stone (from top, taking downward positive): x = ½gt2 = 5t2.
Thrown-up stone (from ground, taking upward positive): height risen y = 20t − ½gt2 = 20t − 5t2.
They meet when the distance fallen plus the height risen equals the tower height: x + y = 45 ⟹ 5t2 + 20t − 5t2 = 45 ⟹ 20t = 45 ⟹ t = 2.25 s.
Distance fallen by dropped stone: x = 5(2.25)2 = 5(5.0625) = 25.3 m, so they meet 25.3 m below the top of the tower (i.e. 45 − 25.3 = 19.7 m above the ground).
Short answer A body travels a distance of 20 m in the 7th second and 24 m in the 9th second. Assuming uniform acceleration, find the initial velocity and the acceleration of the body.Frequently asked3 m
s7 = u + a(6.5) = 20
s9 = u + a(8.5) = 24
Subtracting: a(8.5 − 6.5) = 24 − 20 ⟹ 2a = 4 ⟹ a = 2 m/s2.
Substituting back: u + 2(6.5) = 20 ⟹ u + 13 = 20 ⟹ u = 7 m/s.
Long answer Derive the three equations of motion (v = u + at, s = ut + ½at2, v2 = u2 + 2as) for uniformly accelerated motion using calculus. Then use them to solve: a bullet fired into a fixed target loses half its velocity after penetrating 3 cm. Find how much further it will penetrate before coming to rest, assuming it faces uniform retardation.Frequently asked5 m
Since a = dv/dt is constant, ∫uvdv = ∫0ta·dt ⟹ v − u = at ⟹ v = u + at.
Since v = dx/dt = u + at, ∫0sdx = ∫0t(u+at)dt ⟹ s = ut + ½at2.
Since a = v(dv/dx) (chain rule), a·dx = v·dv, so a∫0sdx = ∫uvv·dv ⟹ as = (v2−u2)/2 ⟹ v2 = u2 + 2as.
Application: Let initial speed = u, after 3 cm speed = u/2. Using v2 = u2 + 2as: (u/2)2 = u2 + 2a(3) ⟹ u2/4 − u2 = 6a ⟹ −3u2/4 = 6a ⟹ a = −u2/8 (retardation).
Now find further penetration distance d, from speed u/2 to 0: 0 = (u/2)2 + 2a·d ⟹ 0 = u2/4 + 2(−u2/8)d ⟹ u2/4 = (u2/4)d ⟹ d = 1 cm.
Long answer State what is meant by relative velocity in one dimension, and derive the expression for the relative velocity of one object with respect to another. Then solve: a jet airplane travelling at 500 km/h ejects its products of combustion at a speed of 1500 km/h relative to the jet plane. Find the speed of the ejected gases relative to the ground observer.Frequently asked5 m
Application: Take the direction of the jet's motion as positive. Jet's velocity vjet = +500 km/h. The exhaust gas moves backward relative to the jet at 1500 km/h, so its velocity relative to the jet, vgas,jet = −1500 km/h.
Velocity of gas relative to ground: vgas = vgas,jet + vjet = −1500 + 500 = −1000 km/h.
So the ejected gas moves at 1000 km/h in the direction opposite to the jet's motion, as seen from the ground.
Long answer A driver sees a red traffic light and applies brakes 0.5 s after seeing it (reaction time). If the car was moving at 54 km/h and decelerates uniformly at 6 m/s2 once the brakes are applied, find (a) the distance travelled during the reaction time, (b) the braking distance after the brakes are applied, (c) the total stopping distance, and explain how reaction time affects stopping distance in general.Frequently asked5 m
(a) During reaction time, the car moves at constant speed (brakes not yet applied): distance = u × tr = 15 × 0.5 = 7.5 m.
(b) After braking, using v2 = u2 + 2as with v = 0, u = 15 m/s, a = −6 m/s2: 0 = (15)2 − 2(6)s ⟹ s = 225/12 = 18.75 m.
(c) Total stopping distance = 7.5 + 18.75 = 26.25 m.
In general, the reaction-time distance grows linearly with speed (u·tr), while the braking distance grows with the square of speed (u2/2a); this is why stopping distance increases sharply, not just proportionally, as speed goes up, which is the physics reason speed limits matter more than they seem to at first glance.
Case-based A car starts from rest at a traffic signal and accelerates uniformly at 3 m/s2 for 8 seconds. It then moves at the constant velocity it has reached for the next 12 seconds, and finally the driver applies brakes, bringing the car to rest with uniform retardation in 4 seconds.
(i) What is the velocity of the car at the end of the acceleration phase?
(ii) What is the distance covered during the acceleration phase?
(iii) What is the distance covered during the constant-velocity phase?
(iv) What is the retardation during braking, and the total distance covered by the car in its entire journey?Frequently asked4 m
(ii) s1 = ut + ½at2 = 0 + ½(3)(64) = 96 m.
(iii) s2 = velocity × time = 24 × 12 = 288 m.
(iv) Retardation: using v = u + at with v = 0, u = 24 m/s, t = 4 s: 0 = 24 + a(4) ⟹ a = −6 m/s2, so retardation = 6 m/s2. Distance in braking phase, s3 = (u+v)/2 × t = (24+0)/2 × 4 = 48 m. Total distance = 96 + 288 + 48 = 432 m.
Case-based Two friends, Rohan and Simran, start from the same point on a straight road at the same instant. Rohan walks at a constant speed of 5 km/h. Simran starts 0.4 hours (24 minutes) later on a bicycle at a constant speed of 15 km/h, along the same road in the same direction.
(i) How far has Rohan walked by the time Simran starts?
(ii) What is the velocity of Simran relative to Rohan once Simran starts moving?
(iii) How long after Simran starts does she catch up with Rohan?
(iv) At what distance from the starting point does Simran catch up with Rohan?Frequently asked4 m
(ii) Both move in the same direction, so relative velocity of Simran with respect to Rohan = 15 − 5 = 10 km/h.
(iii) Simran must close the 2 km gap at a relative speed of 10 km/h: time = distance/relative velocity = 2/10 = 0.2 h = 12 minutes.
(iv) Distance covered by Simran in that time = 15 × 0.2 = 3 km from the starting point (check: Rohan's total time is 0.4 + 0.2 = 0.6 h, distance = 5 × 0.6 = 3 km — matches).
Test yourselfPractice quiz
Marks lost every yearCommon mistakes
- Treating distance and displacement as the same thing, especially in round-trip problems where displacement is zero but distance is not.
- Averaging two speeds directly (arithmetic mean) for a journey covering equal distances at different speeds, instead of using total distance/total time (the harmonic mean).
- Assuming acceleration must be zero when velocity is zero — missing the classic case of the highest point of a vertical throw, where v = 0 but a = −g.
- Mixing sign conventions midway through a free-fall or braking problem (switching what 'positive' means partway through the calculation), which flips the sign of the final answer.
- Reading the slope of an x–t graph as speed only, without noting that a negative slope means motion in the negative direction, not 'the object is slowing down'.
- Forgetting that the kinematic equations (v=u+at, s=ut+½at², v²=u²+2as) apply only to uniform (constant) acceleration, and trying to use them directly when acceleration itself is changing.
- In relative-velocity problems, subtracting speeds when the two bodies move in opposite directions (should add) or adding when they move in the same direction (should subtract).
- Forgetting to convert km/h to m/s (÷3.6, or ×5/18) before substituting into the SI kinematic equations, which is one of the most common numerical slips.
How to prepareStrategy for this chapter
This chapter rewards understanding over memorising formulas, because most marks lost in the school final come from graph-reading and sign-convention slips, not from forgetting the equations. Start by making sure the distance-vs-displacement and average-speed-vs-average-velocity distinctions are automatic — several 1-mark and assertion-reason questions test nothing but this.
Learn the three kinematic equations well enough to derive them two ways: from a v–t graph (area and slope) and by calculus (integrating a = dv/dt and v = dx/dt). Boards and school finals that ask for a 'derivation' give marks for each correct step, so write the full working, not just the final boxed result.
Practice reading x–t, v–t and a–t graphs until you can instantly say what slope and area mean on each — this is the fastest-scoring skill in the chapter once it clicks. For numericals, always fix a sign convention (write it down) before substituting, convert every speed to consistent SI units first, and keep free-fall and relative-velocity problems as two separate mental templates so you don't mix their sign rules. Finally, work through the NCERT exercise and exemplar problems fully at least once — several of the case-based and long-answer questions in school finals are close variants of these.