Motion in a Straight Line Class 11 Physics Notes

Kinematics is the branch of mechanics that describes how bodies move — position, velocity, acceleration — without asking what causes the motion. When a body is small enough that every part of it moves together, it can be treated as a single point, or particle, for the purpose of describing its motion.A particle's position is fixed by its coordinates relative to a chosen origin. As it moves, it traces a path called its trajectory. Distance is the total length of that path; displacement is the straight-line vector from the starting point to the ending point, regardless of the path taken in between.

MCQ Questions

1 Mark Questions

1. 1
   Define a point object (particle) in the study of motion.
2. 2
   Write the SI unit of displacement.
3. 3
   When is the magnitude of displacement equal to the distance travelled?
4. 4
   Can the displacement of a moving body be zero while the distance travelled is non-zero? Give one example.

2 Mark Questions

1. 1
   Distinguish between distance and displacement, giving one example each.
2. 2
   A body travels 40 m due north and then 30 m due east. Find the total distance travelled and the magnitude of its displacement.
3. 3
   A man walks 5 m north, 3 m east and finally 5 m south. Find the magnitude of his displacement and the distance he walked.
4. 4
   Explain, with an example, why the magnitude of displacement is never greater than the distance travelled.

3 Mark Questions

1. 1
   Define position, trajectory and displacement. A particle moves from x = 2 m to x = 8 m and then back to x = 5 m. Find the distance travelled and its displacement.
2. 2
   A cyclist rides one full round of a square park of side 100 m in 12 minutes. Find (i) the distance travelled, (ii) the magnitude of his displacement, and (iii) his average speed in m/s.

Average speed is total distance divided by total time; average velocity is total displacement divided by total time elapsed. The former is a scalar and can never be zero for a body that has actually moved; the latter is a vector and can be exactly zero even after a long, fast journey — any round trip has zero average velocity no matter how far it travelled.

Instantaneous velocity is the limit of average velocity as the time interval shrinks toward zero — the derivative of position with respect to time, v = dx/dt. Its direction is always tangential to the path at that instant, and its magnitude always equals the instantaneous speed (they only differ for average quantities, never for instantaneous ones).

MCQ Questions

1 Mark Questions

1. 1
   Define average speed.
2. 2
   Define average velocity.
3. 3
   Can the average velocity of a moving body over an interval be zero while its average speed is non-zero? Give one example.
4. 4
   What is the average speed of a body that moves at speeds v₁ and v₂ for two equal time intervals?

2 Mark Questions

1. 1
   Derive the expression for the average speed of a body that covers equal distances at speeds v₁ and v₂.
2. 2
   A train travels from station P to station Q at 40 km/h and returns along the same route at 30 km/h. Find its average speed for the round trip.
3. 3
   Distinguish between average speed and average velocity. When do their magnitudes become equal?
4. 4
   A particle moves at 10 m/s for 5 s and then at 20 m/s for 10 s. Find its average speed.

3 Mark Questions

1. 1
   A body covers the first half of a journey at speed v₁ and the second half at speed v₂. Show that its average speed is 2v₁v₂/(v₁ + v₂).
2. 2
   A man walks 2.5 km to a market at 5 km/h and returns at 7.5 km/h. Find (i) his average speed for the round trip and (ii) the magnitude of his average velocity for the whole trip.

Average acceleration is the change in velocity divided by the time taken; instantaneous acceleration is its limiting value as the interval shrinks to zero, a = dv/dt = d²x/dt². For constant acceleration, average and instantaneous acceleration are identical, which is exactly the condition under which the standard three equations of motion apply.

MCQ Questions

1 Mark Questions

1. 1
   Define acceleration.
2. 2
   What is retardation?
3. 3
   A body starts from rest and has a uniform acceleration of 2 m/s². Find its velocity after 4 s.
4. 4
   Can a body have negative acceleration while it is speeding up? Explain in one line.

2 Mark Questions

1. 1
   Distinguish between acceleration and retardation with an example.
2. 2
   A car moving at 72 km/h is brought to rest by applying brakes in 5 s. Find its retardation.
3. 3
   Why is a negative acceleration not always retardation? Explain with an example.
4. 4
   The velocity of a particle changes from 5 m/s to 25 m/s in 4 s. Find its average acceleration.

3 Mark Questions

1. 1
   Define average acceleration and instantaneous acceleration, and show that they are equal when acceleration is constant.
2. 2
   A train slows uniformly from 72 km/h to 36 km/h while covering a distance of 200 m. Find its retardation and the time taken.
3. 3
   A body moving with uniform acceleration covers 20 m in the 1st second and 40 m in the 2nd second. Find its initial velocity and acceleration.

For constant acceleration a, the three kinematic equations follow directly from integrating a = dv/dt and v = ds/dt over time. They connect any three of the five quantities u, v, a, s, t, so most numericals reduce to picking the equation that already contains the three quantities you know.

A separate, frequently tested result gives the distance covered in a specific second (not up to that second): the displacement during the nth second alone, obtained by subtracting the total displacement up to (n−1) seconds from the total up to n seconds.

MCQ Questions

1 Mark Questions

1. 1
   State the three equations of motion for a uniformly accelerated body.
2. 2
   Write the formula for the displacement of a body during the nth second of its motion.
3. 3
   Under what condition does the nth-second formula give the true distance travelled rather than just the displacement?
4. 4
   A body starts from rest and accelerates uniformly at 5 m/s². Find its velocity after 4 s.

2 Mark Questions

1. 1
   Derive the first equation of motion, v = u + at, using the definition of acceleration.
2. 2
   Derive the expression for the displacement of a body in the nth second: sₙ = u + (a/2)(2n − 1).
3. 3
   A body starting from rest moves with a uniform acceleration of 5 m/s². Find its displacement in the 8th second.
4. 4
   A stone is dropped from rest. Taking g = 10 m/s², find the distance it falls in the 4th second.

3 Mark Questions

1. 1
   Derive the three equations of motion for a body moving with uniform acceleration.
2. 2
   A particle starts from rest with a uniform acceleration of 6 m/s². Find the distance covered (i) in the first 5 s and (ii) in the 5th second.
3. 3
   A body starts from rest with uniform acceleration a. Show that the distances covered in successive seconds are in the ratio 1 : 3 : 5 : 7 ... .

Three graphs describe one-dimensional motion completely, and each one's slope or enclosed area hands you the next quantity down (or up) the chain position → velocity → acceleration.

A position–time graph that is vertical would mean infinite velocity, which is physically impossible — if you ever derive a vertical segment, recheck the setup rather than accepting the result. Similarly, a velocity-time graph can never be vertical for a real object, since that would require infinite acceleration.

MCQ Questions

1 Mark Questions

1. 1
   What does the slope of a velocity–time graph represent?
2. 2
   What does the area under an acceleration–time graph give?
3. 3
   What is the nature of the x–t graph of a body at rest?
4. 4
   Can a position–time graph for a real particle ever be vertical? Give a reason.

2 Mark Questions

1. 1
   Derive the relation s = ut + ½at² using the area under a velocity–time graph.
2. 2
   Draw the v–t graph of a body (i) at rest, (ii) moving with uniform velocity, and (iii) moving with uniform acceleration.
3. 3
   The velocity of a particle increases uniformly from 10 m/s to 25 m/s in 5 s. Using the v–t graph, find the distance covered.
4. 4
   What does the slope of an x–t graph and the area under a v–t graph represent?

3 Mark Questions

1. 1
   Draw the x–t, v–t and a–t graphs for a ball thrown vertically upward until it returns to the thrower (air resistance neglected).
2. 2
   From the velocity–time graph of a uniformly accelerated body, derive all three equations of motion.

Near Earth's surface every object accelerates downward at the same rate g (≈9.8 m/s², often rounded to 10 m/s²), regardless of mass, once air resistance is ignored. This is nothing but the standard equations of motion with a replaced by ±g, split into three shapes depending on how the motion starts.

MCQ Questions

1 Mark Questions

1. 1
   Define free fall.
2. 2
   What is the approximate value of acceleration due to gravity on the Earth's surface?
3. 3
   Write the three kinematic relations for a body dropped from rest.
4. 4
   In what ratio do the distances fallen from rest in successive seconds stand?

2 Mark Questions

1. 1
   Derive the expression for the maximum height reached by a body thrown vertically upward with speed u.
2. 2
   A ball is thrown upward with a speed of 30 m/s. Taking g = 10 m/s², find (i) the maximum height reached and (ii) the total time of flight.
3. 3
   Show that the time of ascent equals the time of descent for a body thrown vertically upward (air resistance neglected).
4. 4
   A stone dropped from a tower reaches the ground in 3 s. Taking g = 10 m/s², find the height of the tower and the velocity of impact.

3 Mark Questions

1. 1
   A ball thrown vertically upward returns to its starting point in 6 s. Taking g = 10 m/s², find (i) its initial velocity and (ii) the maximum height reached.
2. 2
   A stone is dropped from the top of a tower 100 m high and, at the same instant, another stone is thrown vertically upward from the foot of the tower at 25 m/s. Taking g = 10 m/s², find when and where the two stones meet.
3. 3
   A body is thrown vertically upward with a speed of 40 m/s. Taking g = 10 m/s², find (i) its height after 2 s, (ii) the total time of flight, and (iii) the velocity with which it passes the thrower's hand on its way down.

The three equations of motion only hold when acceleration is a constant number. The moment acceleration is given as a function of time, velocity, or position instead, those shortcuts break down and you must return to the defining calculus relations.

Going from acceleration back to velocity, or velocity back to position, means integrating rather than differentiating: if a is given as a function of t, integrate once (using the initial velocity as the constant of integration) to get v(t), then integrate again to get x(t). When acceleration is instead given as a function of velocity or position, the chain-rule form a = v(dv/dx) lets you separate variables in terms of x directly, without ever going through t.

MCQ Questions

1 Mark Questions

1. 1
   Write the differential definitions of velocity and acceleration.
2. 2
   Can the equation v = u + at be used when the acceleration varies with time? Give a reason.
3. 3
   Write the relation a = v(dv/dx). When is this form especially useful?
4. 4
   The displacement of a particle is x = 2t² − 3t. Find its velocity at t = 2 s.

2 Mark Questions

1. 1
   For a particle with x = 3 + 8t + 7t² (x in metres, t in seconds), find its velocity and acceleration at any time t.
2. 2
   A particle starts from rest with an acceleration a = 6t (m/s²). Find its velocity and displacement as functions of time.
3. 3
   Explain why the three kinematic equations cannot be applied when the acceleration is not constant.
4. 4
   The velocity of a particle is v = 3t² + 4t + 2 (m/s). Find its acceleration at t = 1 s.

3 Mark Questions

1. 1
   A particle starts from rest at t = 0 with an acceleration a = 3t² + 2t + 2 (m/s²). Find its velocity and displacement at t = 2 s.
2. 2
   A particle has acceleration a = −kv (k constant) and initial velocity u. Find its velocity as a function of time.
3. 3
   A particle moving along the x-axis has acceleration a = −2x (x in metres). If it starts from rest at x = 2 m, find its speed at x = 1 m.

Every description of motion is made from some vantage point, called a frame of reference. Two passengers sitting still inside a moving train are at rest relative to each other but moving relative to someone standing on the platform — neither description is more correct; they simply use different frames.A frame moving with constant velocity (zero acceleration) is an inertial frame, where Newton's laws hold in their simple form. An accelerating frame is a non-inertial frame, where extra pseudo-forces are needed to make Newton's laws balance — that's a topic properly developed in Laws of Motion.

MCQ Questions

1 Mark Questions

1. 1
   Define a frame of reference.
2. 2
   What is an inertial frame of reference?
3. 3
   Why is a pseudo-force required in a non-inertial frame?
4. 4
   Is a frame moving with constant velocity inertial or non-inertial?

2 Mark Questions

1. 1
   Distinguish between inertial and non-inertial frames of reference, giving one example of each.
2. 2
   Two passengers sit face to face inside a train moving at uniform velocity. Explain why they see each other at rest while a person on the platform sees them in motion.
3. 3
   A lift accelerates upward with acceleration a. Find the effective weight of a man of mass m standing in the lift, as observed from the ground frame.
4. 4
   Is the Earth a perfect inertial frame? Give a reason.

3 Mark Questions

1. 1
   Explain with examples why motion is relative — a body at rest in one frame is in motion in another.
2. 2
   A mass m hangs from a spring balance in a lift. The lift accelerates upward at a. Show, from the lift frame and from the ground frame, that the balance reads m(g + a).
3. 3
   State the condition for a frame to be inertial, and show that any frame moving uniformly with respect to an inertial frame is itself inertial.

If two objects A and B move with velocities v_A and v_B measured from the same fixed origin, the velocity of B as seen by an observer riding on A is the vector difference v_BA = v_B − v_A, and symmetrically v_AB = v_A − v_B = −v_BA.Along a single line this is ordinary signed subtraction: fix one direction as positive, assign each velocity a signed value, and subtract. Two bodies moving the same way have a smaller relative speed than either individual speed; two bodies moving toward each other have a relative speed equal to the sum of their speeds — which is why closing speed problems (two trains approaching, a chase, a collision course) almost always start by adding, not subtracting.

MCQ Questions

1 Mark Questions

1. 1
   Define relative velocity.
2. 2
   Write the expression for the relative velocity of A with respect to B when both move along the same straight line.
3. 3
   Two bodies move along the same line in the same direction with speeds 20 m/s and 15 m/s. What is their relative velocity?
4. 4
   When is the relative velocity of two bodies equal to the sum of their individual speeds?

2 Mark Questions

1. 1
   Derive the expression v_BA = v_B − v_A for two bodies moving along a straight line.
2. 2
   Two cars move toward each other with speeds 20 m/s and 30 m/s. Find the rate at which their separation decreases and the reduction in separation in 10 s.
3. 3
   A 120 m long train moving at 10 m/s overtakes an 80 m long train moving at 6 m/s in the same direction. Find the time taken to completely pass it.
4. 4
   A police van moving at 30 km/h fires a bullet at a car speeding away in the same direction at 192 km/h. If the muzzle speed of the bullet is 150 m/s, find the speed with which the bullet hits the car.

3 Mark Questions

1. 1
   A car moving at 10 m/s is 25 m behind a scooter moving at 8 m/s in the same direction. Find the time after which the car overtakes the scooter and the distance the car covers in that time.
2. 2
   A car moving at 20 m/s is 60 m behind a bus moving at 15 m/s in the same direction. Find (i) how long the car takes to catch the bus and (ii) the distance covered by the car before catching it.
3. 3
   Show that for two bodies moving toward each other along a straight line, the time to collision depends only on the initial separation and the sum of their speeds.

The standard application of relative velocity is a boat crossing a flowing river — equally, a swimmer against a current, an aircraft in crosswind, or a person walking through falling rain. The object's velocity relative to the ground is always the vector sum of its velocity relative to the medium and the medium's velocity relative to the ground.Along a straight channel this becomes simple addition or subtraction of speeds: moving with the current adds the current's speed to your own; moving against it subtracts. When crossing perpendicular to a current, resolve the ground velocity into a component along the flow (which causes downstream drift) and a component across it (which determines crossing time).

MCQ Questions

1 Mark Questions

1. 1
   Write the velocity-addition relation for an object moving in a river: velocity relative to the ground in terms of its velocity relative to the water and the current's velocity.
2. 2
   For a swimmer crossing a river perpendicular to the flow, what determines the time of crossing?
3. 3
   What is meant by drift in a river-boat problem?
4. 4
   A boat's speed in still water is u and the current speed is w. What is its speed downstream?

2 Mark Questions

1. 1
   A boat rows perpendicular to the banks of a 1 km wide river at 5 km/h in still water, while the current flows at 3 km/h. Find the time to cross and the drift.
2. 2
   Rain falls vertically at 3 m/s and a man runs horizontally at 4 m/s. Find the speed with which the rain strikes the man and the direction in which he should hold his umbrella.
3. 3
   Derive an expression for the drift of a swimmer who aims perpendicular to a river of width d, with current speed w and still-water speed u.
4. 4
   A boy rows at 4 m/s in still water across a river 400 m wide flowing at 3 m/s, aiming perpendicular to the flow. Find how far downstream he lands and his resultant speed relative to the ground.

3 Mark Questions

1. 1
   Derive the condition for a swimmer to cross a river along the shortest possible path: the angle at which he must aim and the time taken, in terms of u, w and the width d.
2. 2
   A swimmer crosses a 200 m wide river at 4 m/s in still water aiming perpendicular to the flow, while the current is 3 m/s. Compute (i) the time to cross, (ii) the point of landing downstream, and (iii) the angle the actual path makes with the straight-across direction.
3. 3
   Rain is falling with a velocity of 10 m/s at 30° to the vertical, and a man is running horizontally at 5 m/s in the direction in which the rain has its horizontal component. Find the velocity of the rain relative to the man.

Exercises (NCERT Q2.1–Q2.18)

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.

2.2) The position–time (x–t) graphs for two children A and B returning from school O to their homes P and Q are shown in Fig. 2.9. Choose the correct entries: (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).

2.3) A woman starts from her home at 9.00 am, walks at 5 km/h on a straight road to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by auto at 25 km/h. Choose suitable scales and plot the x–t graph of her motion.

2.4) A 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 and determine how long he takes to fall into a pit 13 m from the start.

2.5) A car moving along a straight highway at 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?

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? (c) Taking x = 0 m and t = 0 s at the highest point with the downward direction positive, give the signs of position, velocity and acceleration during the upward and downward motion. (d) To what height does the ball rise and after how long does it return to the player's hands? (Take g = 9.8 m/s² and neglect air resistance.)

2.7) Read each statement below carefully and state with reasons and examples whether 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 a positive value of acceleration must be speeding up.

2.8) A 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 and t = 12 s.

2.9) Explain clearly, with examples, the distinction between (a) the magnitude of displacement over an interval of time and the total length of the path covered by the particle over the same interval, and (b) the magnitude of average velocity over an interval and the average speed over the same interval. Show in both cases that the second quantity is either greater than or equal to the first. When is the equality sign true?

2.10) A man walks on a straight road from his home to a market 2.5 km away at a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home at 7.5 km/h. What is the (a) magnitude of average velocity, and (b) average speed of the man over the intervals (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min?

2.11) In Exercises 2.9 and 2.10 we carefully distinguished between average speed and the magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and the magnitude of instantaneous velocity. Why?

2.12) Look 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.

2.13) Figure 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.

2.14) A police van moving on a highway at 30 km/h fires a bullet at a thief's car speeding away in the same direction at 192 km/h. If the muzzle speed of the bullet is 150 m/s, with what speed does the bullet hit the thief's car? (Obtain the speed relevant for damaging the car.)

2.15) Suggest a suitable physical situation for each of the following graphs (Fig. 2.12): (a) an x–t graph that rises and then suddenly drops back to zero, (b) a v–t graph of alternating positive and negative spikes that gradually die out, (c) an a–t graph that is zero except for a short pulse.

2.16) Figure 2.13 gives the x–t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration of the particle at t = 0.3 s, 1.2 s and −1.2 s.

2.17) Figure 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 the average velocity for each interval.

2.18) Figure 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?

Answers at a glance

1. 1
   Q2.1 — (a), (b) Yes; (c), (d) No.
2. 2
   Q2.2 — (a) A; (b) A; (c) B; (d) same time; (e) B, once.
3. 3
   Q2.3 — Three segments: slope +5 km/h from (0,0) to (0.5 h, 2.5 km), horizontal at 2.5 km from 0.5 h to 8.0 h, slope −25 km/h back to (8.1 h, 0); she reaches home at about 5.06 pm.
4. 4
   Q2.4 — 37 s.
5. 5
   Q2.5 — Retardation ≈ 3.06 m/s²; time to stop ≈ 11.4 s.
6. 6
   Q2.6 — (a) Vertically downward (magnitude g); (b) v = 0, a = g downward; (c) upward: x<0, v<0, a>0; downward: x>0, v>0, a>0; (d) 44.1 m; 6 s.
7. 7
   Q2.7 — (a) True; (b) False; (c) True; (d) False.
8. 8
   Q2.8 — Saw-tooth speed–time graph: speed grows with slope g to 42 m/s at t ≈ 4.3 s, drops to 37.8 m/s at the collision, decays and returns to 37.8 m/s at about t = 12.0 s before the next collision.
9. 9
   Q2.9 — Path length ≥ magnitude of displacement, and average speed ≥ magnitude of average velocity; equality holds only when the motion is unidirectional (no reversal of direction).
10. 10
    Q2.10 — (i) 5 km/h and 5 km/h; (ii) 0 and 6 km/h; (iii) 1.875 km/h and 5.625 km/h.
11. 11
    Q2.11 — Over an infinitesimal interval the particle cannot reverse direction, so the path length equals the magnitude of the displacement; dividing by the interval makes speed equal to the magnitude of velocity.
12. 12
    Q2.12 — None of the four graphs (a)–(d) can represent one-dimensional motion: two values of x or v at a single instant (a), (b), a negative speed (c), and a decreasing total path length (d) are all impossible.
13. 13
    Q2.13 — No. The graph is an x–t plot, not the spatial path; in one-dimensional motion the particle always moves along a straight line. Suitable context: a ball held at rest until t = 0 and then released to fall freely (x ∝ t²).
14. 14
    Q2.14 — 105 m/s (the speed of the bullet relative to the thief's car).
15. 15
    Q2.15 — (a) A ball thrown vertically upward, caught on its return; (b) a ball bouncing repeatedly on a hard floor with diminishing rebounds; (c) a cricket ball moving uniformly being hit by a bat — a short, large acceleration during impact.
16. 16
    Q2.16 — t = 0.3 s: x<0, v<0, a>0; t = 1.2 s: x>0, v>0, a<0; t = −1.2 s: x<0, v>0, a>0.
17. 17
    Q2.17 — Average speed is greatest in interval 3 and least in interval 2; average velocity is positive in intervals 1 and 2 and negative in interval 3.
18. 18
    Q2.18 — Average acceleration is greatest in magnitude in interval 2, average speed is greatest in interval 3; v is positive in all three intervals, a is positive in interval 1, negative in interval 2 and positive in interval 3; accelerations at A, B, C and D are all zero.

MCQ Questions

1 Mark Questions

1. 1
   Define average velocity.
2. 2
   State the three equations of motion for a uniformly accelerated body.
3. 3
   What does the slope of a velocity–time graph represent?
4. 4
   Define free fall.
5. 5
   When does the formula sₙ = u + (a/2)(2n − 1) give the true distance travelled in the nth second?

2 Mark Questions

1. 1
   A train moving at 54 km/h is brought to rest by applying brakes uniformly in 18 s. Find the acceleration and the distance covered before stopping.
2. 2
   Distinguish between the average speed and the average velocity of a body that returns to its starting point.
3. 3
   A stone is dropped from the top of a tower 45 m high. Taking g = 10 m/s², find the time taken to reach the ground and the velocity of impact.
4. 4
   A swimmer aims perpendicular to a 150 m wide river at 3 m/s in still water, while the current flows at 4 m/s. Find the time to cross and the drift.

3 Mark Questions

1. 1
   A balloon is ascending at 5 m/s at a height of 300 m above the ground when a stone is dropped from it. Taking g = 10 m/s², find the time taken by the stone to reach the ground and the speed with which it hits the ground.
2. 2
   Balls are released from the top of a tower at intervals of 1 s. When the 4th ball is just released, find the separations between the 1st and 2nd balls and between the 2nd and 3rd balls. (g = 10 m/s²)
3. 3
   A body initially at rest accelerates uniformly at 2 m/s² for 5 s, then moves with constant velocity for 4 s, and finally retards uniformly to rest in 3 s. Draw the velocity–time graph of the motion and find the total distance covered.

5 Mark Questions

1. 1
   (a) Derive the three equations of motion for a body moving with uniform acceleration using calculus. (b) A car moving at 72 km/h is brought to rest with uniform retardation in 5 s. Find the retardation and the distance covered.
2. 2
   (a) What do the slope and the area under a velocity–time graph represent? (b) Draw the x–t, v–t and a–t graphs for a ball thrown vertically upward until it returns to the thrower. (c) From the v–t graph, obtain the maximum height and the total time of flight for an initial speed u.