NEET Physics: Units, Measurement and Kinematics · Chapter 2

Motion in a Straight Line

What to remember

Displacement is the shortest straight-line change in position (a vector). Path length is the total distance travelled (a scalar). Path length is always greater than or equal to the magnitude of displacement.
Velocity is the slope of the position-time graph; acceleration is the slope of the velocity-time graph. The area under a velocity-time graph is displacement; the area under an acceleration-time graph is the change in velocity.
For uniform acceleration: v = u + at, s = ut + ½at², v² = u² + 2as. Distance travelled in the nth second is s_n = u + a(2n − 1)/2.
Free fall is uniformly accelerated motion with a = g = 9.8 m s⁻² downward. Take upward as positive: a = −g. At the highest point, v = 0 but a = −g.

Position, path length and displacement

Motion is the change of position with time. To describe motion along a straight line, we choose an origin and a positive direction. Positions to one side are positive and to the other side are negative. This line is called the x-axis.

Path length (distance) is the total length of the path travelled. It is a scalar. It is never negative and never decreases with time.
Displacement is the change in position: Δx = x₂ − x₁. It is a vector; in one dimension, its sign gives its direction. It can be positive, negative or zero.
The magnitude of displacement is less than or equal to the path length. They are equal only when the object moves in a straight line without changing direction.

Example: A man walks 6 m east and then 2 m west. Path length = 8 m. Displacement = 6 − 2 = 4 m east.

Average and instantaneous speed and velocity

Quantity	Definition	Scalar or vector	Unit
Average speed	total path length / total time	scalar	m s⁻¹
Average velocity	displacement / time interval = Δx/Δt	vector	m s⁻¹
Instantaneous velocity	v = lim (Δt → 0) Δx/Δt = dx/dt	vector	m s⁻¹
Instantaneous speed	magnitude of instantaneous velocity	scalar	m s⁻¹

Facts:

Average speed is always greater than or equal to the magnitude of average velocity.
For a round trip, the average velocity is zero but the average speed is not zero.
If an object covers equal distances at speeds v₁ and v₂, the average speed is the harmonic mean 2v₁v₂/(v₁ + v₂), not the simple mean.
If an object moves for equal times at speeds v₁ and v₂, the average speed is (v₁ + v₂)/2.
The instantaneous speed is the magnitude of instantaneous velocity. This is true at each instant; but average speed is not the magnitude of average velocity in general.

Acceleration

Average acceleration: a_avg = Δv/Δt = (v₂ − v₁)/(t₂ − t₁). Unit: m s⁻². Dimensions [L T⁻²].
Instantaneous acceleration: a = dv/dt = d²x/dt².
Uniform acceleration means constant acceleration; the instantaneous and average accelerations are then equal.
Retardation (deceleration) means acceleration opposite to the velocity. The speed decreases. A negative acceleration is not always retardation: if velocity is also negative, speed increases.
Motion with zero acceleration is uniform motion: the velocity is constant, and the position-time graph is a straight line.

Graphs of motion (described in words)

Graph	Shape for uniform velocity	Shape for uniform acceleration	Slope means	Area means
Position-time (x-t)	straight inclined line	parabola	velocity	no standard meaning
Velocity-time (v-t)	horizontal line	straight inclined line	acceleration	displacement
Acceleration-time (a-t)	line on the time axis (a = 0)	horizontal line	rate of change of acceleration	change in velocity

More facts about graphs:

A steeper x-t line means a larger speed. A horizontal x-t line means the object is at rest.
On an x-t graph, if the curve bends upward (concave up), the acceleration is positive. If it bends downward, the acceleration is negative.
On a v-t graph, area above the time axis is positive displacement and area below is negative displacement. The sum with signs gives displacement. The sum of areas without signs gives path length.
A v-t line cutting the time axis means the object momentarily stops and reverses its direction.
The slope of the chord on an x-t graph gives the average velocity. The slope of the tangent gives the instantaneous velocity.

Elementary calculus used in motion

Differentiation gives rates of change: v = dx/dt and a = dv/dt. Integration is the reverse process.

Basic rules used: d(tⁿ)/dt = n tⁿ⁻¹ and ∫tⁿ dt = tⁿ⁺¹/(n + 1) (n ≠ −1).

Example: If x = 5 + 3t² (x in m, t in s), then v = dx/dt = 6t and a = dv/dt = 6 m s⁻². At t = 2 s, v = 12 m s⁻¹.

Example: If a = 4t, with u = 0 at t = 0, then v = ∫a dt = 2t². At t = 3 s, v = 18 m s⁻¹. Displacement from 0 to 3 s = ∫2t² dt = (2/3)t³ = 18 m.

Another useful relation: a = dv/dt = v dv/dx. It is used when acceleration is given as a function of position.

Kinematic equations for uniformly accelerated motion

Derivation by graph. Draw the v-t graph for an object with initial velocity u and acceleration a: a straight line from u at t = 0 to v at time t. Its slope is a = (v − u)/t, so v = u + at. The area under the line is a trapezium: s = ½(u + v)t. Putting v = u + at gives s = ut + ½at². Eliminating t between v = u + at and s = ½(u + v)t gives v² = u² + 2as.

Derivation by calculus. a = dv/dt is constant. Integrating, v = u + at. Then v = dx/dt = u + at; integrating, x = x₀ + ut + ½at². With a = v dv/dx, integrating gives v² = u² + 2a(x − x₀).

Equation	Missing variable
v = u + at	s
s = ut + ½at²	v
v² = u² + 2as	t
s = ½(u + v)t	a

Here u is the initial velocity, v the final velocity, a the constant acceleration, t the time, s the displacement. Use the sign convention: choose a positive direction; velocities, accelerations and displacements in the other direction are negative.

Distance in the nth second. The displacement in the nth second is the displacement in n seconds minus that in (n − 1) seconds:

s_n = [un + ½an²] − [u(n − 1) + ½a(n − 1)²] = u + a(2n − 1)/2.

Check: for a body starting from rest, s₁ : s₂ : s₃ = 1 : 3 : 5 (distances in successive equal seconds).

Starting from rest, the distances covered in total times 1 s, 2 s, 3 s are in the ratio 1 : 4 : 9.

Stopping distance: a vehicle with initial speed u braking with retardation a stops in distance u²/(2a). The stopping distance is proportional to u². Doubling the speed multiplies the stopping distance by four.

Free fall and motion under gravity

Free fall is motion under gravity alone, with air resistance neglected. Near the earth's surface, g = 9.8 m s⁻² downward, the same for all bodies. In numerical questions, g = 10 m s⁻² is often given for simplicity.

Take the upward direction as positive, so a = −g.

Case	Formulae
Dropped from rest from height h	time t = √(2h/g); speed on landing v = √(2gh)
Thrown upward with speed u	maximum height H = u²/(2g); time to top = u/g; total time of flight (return to the same level) = 2u/g
Thrown downward with speed u from height h	v = u + gt; h = ut + ½gt²; v² = u² + 2gh
Speed on return to the starting point	equal to u (opposite direction)

Facts:

At the highest point, velocity is zero but acceleration is still g downward.
For a body thrown upward, the time of ascent equals the time of descent to the same level, and the speed at any height is the same going up and coming down.
A stone dropped from a height falls distances in the ratio 1 : 3 : 5 in successive seconds.
Dropped from a balloon: if a balloon is rising with velocity u and a stone is released from it, the stone has initial velocity u upward at the moment of release (the stone first rises, then falls). If the balloon is at rest, the stone is simply dropped. If the balloon is descending with velocity u, the stone starts with u downward.

Worked example 1. A ball is thrown upward with 20 m s⁻¹ (take g = 10 m s⁻²). Maximum height = 20²/(2 × 10) = 20 m. Time to top = 20/10 = 2 s. Total time to return = 4 s.

Worked example 2. A stone is dropped from a tower 80 m high (g = 10 m s⁻²). Time = √(2 × 80/10) = 4 s. Speed on landing = 10 × 4 = 40 m s⁻¹. Distance fallen in the last second = 80 − ½ × 10 × 3² = 80 − 45 = 35 m.

Worked example 3. A body is released from a balloon rising at 10 m s⁻¹ when the balloon is 75 m above the ground (g = 10 m s⁻²). Taking upward as positive: −75 = 10t − 5t². So t² − 2t − 15 = 0, giving t = 5 s. The speed on landing is |10 − 10 × 5| = 40 m s⁻¹.

Relative velocity in one dimension

The velocity of object A relative to object B is v_AB = v_A − v_B, where both velocities are measured in the same frame and the same positive direction.

If A and B move in the same direction, the relative speed is the difference of the speeds.
If they move in opposite directions, the relative speed is the sum of the speeds.
v_BA = −v_AB.
For two objects approaching each other along a line, the time to meet = initial separation / relative speed.

Example: Two trains A and B move on parallel tracks at 90 km h⁻¹ and 60 km h⁻¹ in the same direction. The relative speed of A with respect to B is 30 km h⁻¹ = 25/3 m s⁻¹.

For uniformly accelerated objects, the relative acceleration is a_AB = a_A − a_B. If the relative acceleration is zero, the relative velocity remains constant.

Formula summary

Item	Formula	Symbols and units
Average velocity	Δx/Δt	m s⁻¹
Instantaneous velocity	dx/dt	m s⁻¹
Acceleration	dv/dt = v dv/dx	m s⁻²
First equation	v = u + at	u, v in m s⁻¹; a in m s⁻²; t in s
Second equation	s = ut + ½at²	s in m
Third equation	v² = u² + 2as
nth second	s_n = u + a(2n − 1)/2	m
Free fall from rest	v = √(2gh), t = √(2h/g)	g = 9.8 m s⁻²
Maximum height	H = u²/(2g)	m
Relative velocity	v_AB = v_A − v_B	m s⁻¹

NEET traps

1. Average speed is not the magnitude of average velocity. They are equal only for straight-line motion without reversal.
2. Equal distances at two speeds: use the harmonic mean 2v₁v₂/(v₁ + v₂), not (v₁ + v₂)/2.
3. At the top of its flight a ball has zero velocity but non-zero acceleration (g).
4. A negative acceleration does not always mean retardation; check the sign of the velocity.
5. The equations v = u + at and others hold only for constant acceleration.
6. Distance in the nth second is a "per second" quantity, u + a(2n − 1)/2, not the distance in n seconds.
7. The area under a v-t graph is displacement, not distance, unless the velocity stays positive.
8. The slope of an x-t graph is velocity, but the slope of a v-t graph is acceleration.
9. A stone dropped from a rising balloon first moves upward with the balloon's velocity.
10. The stopping distance grows as the square of the initial speed, not in proportion to it.

One-liners

1. Displacement can be zero while path length is not.
2. Instantaneous velocity is the slope of the tangent to the x-t graph.
3. The area under an a-t graph gives the change in velocity.
4. A straight inclined x-t line means uniform velocity; a parabolic x-t line means uniform acceleration.
5. For constant acceleration, the average velocity is (u + v)/2.
6. Starting from rest, distances in successive equal seconds are in the ratio 1 : 3 : 5 : 7.
7. Starting from rest, total distances in 1 s, 2 s, 3 s are in the ratio 1 : 4 : 9.
8. A body thrown up with speed u rises to u²/(2g).
9. The time of flight of a body thrown up and returning to the starting level is 2u/g.
10. A body dropped from height h reaches the ground with speed √(2gh).
11. For a round trip, average velocity is zero.
12. a = v dv/dx is used when acceleration depends on position.
13. In free fall, all bodies fall with the same acceleration when air resistance is neglected.
14. The relative velocity of A with respect to B is v_A − v_B.
15. For objects moving in opposite directions, the relative speed is the sum of the speeds.

Practice questions

A man walks 12 m east and then 5 m west along a straight road. His path length and the magnitude of his displacement are respectively
1. 17 m and 17 m
2. 7 m and 7 m
3. 17 m and 7 m
4. 7 m and 17 m
Answer
C. 17 m and 7 m
Path length = 12 + 5 = 17 m. Displacement = 12 − 5 = 7 m.
A car travels the first half of a distance at 40 km h⁻¹ and the second half at 60 km h⁻¹. Its average speed for the whole journey is
1. 24 km h⁻¹
2. 48 km h⁻¹
3. 52 km h⁻¹
4. 50 km h⁻¹
Answer
B. 48 km h⁻¹
For equal distances, v_avg = 2v₁v₂/(v₁ + v₂) = 2 × 40 × 60/100 = 48 km h⁻¹ (not the simple mean 50).
The position of a particle is x = 2t³ − 3t² (x in metre, t in second). Its velocity at t = 2 s is
1. 12 m s⁻¹
2. 4 m s⁻¹
3. 16 m s⁻¹
4. 24 m s⁻¹
Answer
A. 12 m s⁻¹
v = dx/dt = 6t² − 6t = 24 − 12 = 12 m s⁻¹.
The area under the velocity-time graph of a particle moving in a straight line between two instants gives its
1. jerk
2. average speed
3. acceleration
4. displacement
Answer
D. displacement
Area under v-t is ∫v dt, which is the displacement (with sign).
A body starts from rest with a uniform acceleration of 2 m s⁻². The distance it travels in the 5th second is
1. 9 m
2. 11 m
3. 10 m
4. 25 m
Answer
A. 9 m
s_n = u + a(2n − 1)/2 = 0 + 2 × 9/2 = 9 m. Check: 25 − 16 = 9 m.
A car moving at 20 m s⁻¹ is brought to rest by a uniform retardation of 5 m s⁻². The distance covered before stopping is
1. 80 m
2. 40 m
3. 4 m
4. 20 m
Answer
B. 40 m
s = u²/(2a) = 400/10 = 40 m.
A stone is dropped from a height of 45 m. Taking g = 10 m s⁻², the time taken to reach the ground is
1. 2 s
2. 4.5 s
3. 9 s
4. 3 s
Answer
D. 3 s
t = √(2h/g) = √(90/10) = 3 s; check: ½ × 10 × 9 = 45 m.
A ball is thrown vertically upward with a speed of 30 m s⁻¹. Taking g = 10 m s⁻², the maximum height reached is
1. 15 m
2. 90 m
3. 45 m
4. 30 m
Answer
C. 45 m
H = u²/(2g) = 900/20 = 45 m.
The velocity of a particle is v = 4t − t² (SI units). Its acceleration at t = 1 s is
1. 4 m s⁻²
2. 2 m s⁻²
3. 3 m s⁻²
4. 0
Answer
B. 2 m s⁻²
a = dv/dt = 4 − 2t = 2 m s⁻² at t = 1 s.
Two cars are 500 m apart and move towards each other along a straight road with speeds 20 m s⁻¹ and 30 m s⁻¹. They meet after
1. 20 s
2. 25 s
3. 10 s
4. 50 s
Answer
C. 10 s
Relative speed = 20 + 30 = 50 m s⁻¹; time = 500/50 = 10 s.
A body starts from rest with uniform acceleration. The distances covered in the first, second and third seconds are in the ratio
1. 1 : 4 : 9
2. 1 : 5 : 9
3. 1 : 2 : 3
4. 1 : 3 : 5
Answer
D. 1 : 3 : 5
s_n ∝ (2n − 1), giving 1 : 3 : 5.
In a position-time graph of a particle, a straight line parallel to the time axis indicates that the particle is
1. at rest
2. moving with uniform acceleration
3. moving with uniform velocity
4. moving with uniform retardation
Answer
A. at rest
Position does not change with time, so the velocity is zero.
A ball is thrown upward. At its highest point,
1. velocity is zero and acceleration is g downward
2. velocity is zero and acceleration is zero
3. velocity is g and acceleration is zero
4. velocity and acceleration are both upward
Answer
A. velocity is zero and acceleration is g downward
Only the gravitational pull acts, so a = g downward at every instant, including the top.
A car increases its speed uniformly from 10 m s⁻¹ to 20 m s⁻¹ while covering 150 m. Its acceleration is
1. 3 m s⁻²
2. 0.5 m s⁻²
3. 2 m s⁻²
4. 1 m s⁻²
Answer
D. 1 m s⁻²
a = (v² − u²)/(2s) = (400 − 100)/300 = 1 m s⁻². Check: t = 150/15 = 10 s, a = 10/10 = 1.
A stone is thrown vertically downward with a speed of 5 m s⁻¹ from the top of a 100 m cliff. Taking g = 10 m s⁻², it reaches the ground after
1. 5 s
2. 4 s
3. 3 s
4. 2 s
Answer
B. 4 s
100 = 5t + 5t², so t² + t − 20 = 0, giving t = 4 s. Check: 5 × 4 + 5 × 16 = 100.
The position of a particle is x = 3 + 4t − t² (SI units). The particle is momentarily at rest at t =
1. 1 s
2. 3 s
3. 2 s
4. 4 s
Answer
C. 2 s
v = 4 − 2t = 0 gives t = 2 s.
Given below are two statements: Statement I: The average speed of a particle can be greater than the magnitude of its average velocity. Statement II: For a round trip the average velocity is zero.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Average speed = path/time ≥ |displacement|/time. For a round trip, displacement is zero. Both are correct.
Given below are two statements: Statement I: The displacement of a body can be zero even when its path length is not zero. Statement II: The path length of a body can be zero even when its displacement is not zero.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
If the path length is zero the body has not moved, so displacement is also zero. Statement II is incorrect.
Given below are two statements: Statement I: The slope of the position-time graph gives the acceleration. Statement II: The area under the velocity-time graph gives the displacement.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
The slope of the position-time graph gives velocity. Statement I is incorrect; Statement II is correct.
Given below are two statements: Statement I: At the highest point of a ball thrown upward, its acceleration is zero. Statement II: For a ball thrown upward, the time of ascent is less than the time of descent to the starting level.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Acceleration is g at the top, and the times are equal in the absence of air resistance. Both are incorrect.
Given below are two statements: Statement I: For uniform acceleration, the distance in the nth second is u + a(2n − 1)/2. Statement II: A body starting from rest covers distances in successive seconds in the ratio 1 : 2 : 3.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
The nth-second formula is correct, but the ratio is 1 : 3 : 5. Statement II is incorrect.
Which of the following statements are correct? 1. If a body moves for equal times at speeds v₁ and v₂, the average speed is (v₁ + v₂)/2. 2. If a body covers equal distances at speeds v₁ and v₂, the average speed is (v₁ + v₂)/2. 3. The stopping distance of a vehicle is proportional to the square of its initial speed.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
Statement 2 is wrong: for equal distances the average speed is 2v₁v₂/(v₁ + v₂). Statements 1 and 3 are correct.
Which of the following statements are correct? 1. Velocity is dx/dt. 2. Acceleration is dv/dt. 3. Acceleration can be written as v dv/dx.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
D. 1, 2 and 3
All three are correct; a = dv/dt = (dv/dx)(dx/dt) = v dv/dx.
Which of the following statements are correct? 1. A stone released from a rising balloon starts with the balloon's upward velocity. 2. The time of ascent equals the time of descent for a ball thrown up and caught at the same level, when air resistance is neglected. 3. At the highest point the velocity is zero, so the acceleration is also zero.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 2 only
Statements 1 and 2 are correct. At the top the acceleration is still g, so 3 is wrong.
Given below are two statements: Statement I: In vacuum, a feather and a stone dropped from the same height reach the ground together. Statement II: All freely falling bodies near the earth have the same acceleration when air resistance is neglected.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are correct and the second explains the first.
Given below are two statements: Statement I: The instantaneous speed of a particle always equals its average speed. Statement II: The average velocity over any motion with changing acceleration is always (u + v)/2.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Instantaneous speed changes in general, and (u + v)/2 holds only for constant acceleration. Both are incorrect.
Assertion (A): At the highest point of a ball thrown upward, the velocity is zero but the acceleration is not zero. Reason (R): The gravitational force acts on the ball throughout its motion.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Gravity gives a constant acceleration g, so v = 0 at the top does not make a = 0.
Assertion (A): The average speed of a particle is never less than the magnitude of its average velocity. Reason (R): The path length is never less than the magnitude of the displacement.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Divide both path length and displacement by the same time; the inequality carries over.
Assertion (A): When the speed of a car is doubled, its stopping distance becomes four times for the same retardation. Reason (R): The stopping distance is directly proportional to the speed.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
s = u²/(2a), so s ∝ u². A is true; R is false.
Assertion (A): A body cannot have zero velocity and non-zero acceleration at the same instant. Reason (R): Acceleration is the rate of change of velocity.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
A ball at the top of its flight has v = 0 and a = g. A is false; R is true.
Assertion (A): The position-time graph of a uniformly accelerated particle is a straight line. Reason (R): The velocity of a uniformly accelerated particle changes uniformly with time.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
With x = x₀ + ut + ½at² the graph is a parabola. A is false; R is true.
Assertion (A): A ball thrown vertically upward returns to the starting point with the same speed as its speed of projection (air resistance neglected). Reason (R): The velocity of the ball is zero at the highest point.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true. The equal speed follows from v² = u² − 2gh with h = 0 (constant acceleration), not from the velocity being zero at the top.
Match List I with List II. List I: a. Slope of position-time graph b. Slope of velocity-time graph c. Area under velocity-time graph d. Area under acceleration-time graph List II: i. Change in velocity ii. Velocity iii. Displacement iv. Acceleration
1. a-iv, b-ii, c-iii, d-i
2. a-ii, b-iv, c-iii, d-i
3. a-ii, b-iv, c-i, d-iii
4. a-iii, b-iv, c-ii, d-i
Answer
B. a-ii, b-iv, c-iii, d-i
a-ii (velocity), b-iv (acceleration), c-iii (displacement), d-i (change in velocity).
Match List I with List II. A ball is thrown upward with a speed of 20 m s⁻¹ (g = 10 m s⁻²). List I: a. Maximum height b. Time to reach the top c. Total time of flight to return to the starting point d. Speed on returning to the starting point List II: i. 2 s ii. 20 m iii. 4 s iv. 20 m s⁻¹
1. a-ii, b-iii, c-i, d-iv
2. a-ii, b-i, c-iv, d-iii
3. a-i, b-ii, c-iii, d-iv
4. a-ii, b-i, c-iii, d-iv
Answer
D. a-ii, b-i, c-iii, d-iv
H = 400/20 = 20 m; t = 20/10 = 2 s; total 4 s; speed on return = 20 m s⁻¹.
Match List I with List II. List I: a. x = 5 m b. x = 3t c. x = 2t² d. x = 4t − t² List II: i. Uniform acceleration from rest ii. Particle at rest iii. Uniform velocity iv. Initially moving forward, then reverses
1. a-i, b-iii, c-ii, d-iv
2. a-ii, b-iii, c-iv, d-i
3. a-ii, b-iii, c-i, d-iv
4. a-iii, b-ii, c-i, d-iv
Answer
C. a-ii, b-iii, c-i, d-iv
a: constant position (rest); b: v = 3 constant; c: v = 4t from rest; d: v = 4 − 2t, which changes sign at t = 2 s.
Match List I with List II. A body starts from rest with acceleration 2 m s⁻². List I: a. Distance in the 1st second b. Distance in the 3rd second c. Distance in the 4th second d. Distance in the 5th second List II: i. 5 m ii. 1 m iii. 9 m iv. 7 m
1. a-iv, b-i, c-ii, d-iii
2. a-ii, b-i, c-iv, d-iii
3. a-i, b-ii, c-iv, d-iii
4. a-ii, b-i, c-iii, d-iv
Answer
B. a-ii, b-i, c-iv, d-iii
s_n = (2n − 1) m: 1 m, 5 m, 7 m, 9 m.
Match List I with List II. List I: a. v = u + at b. s = ut + ½at² c. v² = u² + 2as d. s = ½(u + v)t List II: i. does not contain acceleration ii. does not contain time iii. does not contain displacement iv. does not contain final velocity
1. a-iii, b-iv, c-ii, d-i
2. a-iv, b-iii, c-ii, d-i
3. a-iii, b-iv, c-i, d-ii
4. a-ii, b-iv, c-iii, d-i
Answer
A. a-iii, b-iv, c-ii, d-i
a-iii, b-iv, c-ii, d-i.
Match List I with List II (speed of B relative to A). List I: a. A at 20 m s⁻¹ east, B at 30 m s⁻¹ east b. A at 20 m s⁻¹ east, B at 30 m s⁻¹ west c. A at 25 m s⁻¹ east, B at 25 m s⁻¹ east d. A at 15 m s⁻¹ west, B at 5 m s⁻¹ east List II: i. 0 ii. 10 m s⁻¹ iii. 50 m s⁻¹ iv. 20 m s⁻¹
1. a-iii, b-ii, c-i, d-iv
2. a-ii, b-iii, c-i, d-iv
3. a-ii, b-i, c-iii, d-iv
4. a-ii, b-iii, c-iv, d-i
Answer
B. a-ii, b-iii, c-i, d-iv
Taking east as positive: +10; −50 (speed 50); 0; +5 − (−15) = 20.
A stone is dropped from a tower and takes 5 s to reach the ground (g = 10 m s⁻²). The distance it covers in the last second is
1. 50 m
2. 80 m
3. 125 m
4. 45 m
Answer
D. 45 m
Height = ½ × 10 × 25 = 125 m. In 4 s it falls 80 m, so the last second gives 125 − 80 = 45 m (or s₅ = 10 × 9/2 = 45 m).
A balloon rises at 5 m s⁻¹. A stone is released from it when it is 60 m above the ground (g = 10 m s⁻²). The stone reaches the ground after
1. √12 s
2. 5 s
3. 4 s
4. 3 s
Answer
C. 4 s
Upward positive: −60 = 5t − 5t², so t² − t − 12 = 0 and t = 4 s. Check: 5(4) − 5(16) = −60.
A ball is thrown vertically upward with 20 m s⁻¹ from the top of a 25 m tower (g = 10 m s⁻²). It reaches the ground after
1. 6 s
2. 5 s
3. 4 s
4. 3 s
Answer
B. 5 s
−25 = 20t − 5t², so t² − 4t − 5 = 0 and t = 5 s. Check: it rises to 20 m above the top in 2 s, then falls 45 m in 3 s: total 5 s.
A ball is dropped from the top of a 100 m tower. At the same instant another ball is thrown upward from the ground with 50 m s⁻¹ (g = 10 m s⁻²). They meet after
1. 2.5 s
2. 1 s
3. 4 s
4. 2 s
Answer
D. 2 s
The relative acceleration is zero, so the relative speed stays 50 m s⁻¹. Time = 100/50 = 2 s. Check: fallen 20 m and risen 80 m, total 100 m.
A car travelling at 72 km h⁻¹ is braked with a uniform retardation of 4 m s⁻². The distance it travels before stopping is
1. 50 m
2. 25 m
3. 100 m
4. 18 m
Answer
A. 50 m
u = 72 km h⁻¹ = 20 m s⁻¹. s = u²/(2a) = 400/8 = 50 m. Check: t = 5 s, average speed 10 m s⁻¹, distance 50 m.
A particle starts with velocity 2 m s⁻¹ along the x-axis. Its acceleration is a = 6t − 2 (SI units). The displacement in the first 2 s is
1. 4 m
2. 12 m
3. 8 m
4. 10 m
Answer
C. 8 m
v = 2 + 3t² − 2t; x = 2t + t³ − t² = 4 + 8 − 4 = 8 m.
A particle moves with velocity v = 10 − 5t (SI units) for 4 s. The total path length it covers in these 4 s is
1. 10 m
2. 20 m
3. 40 m
4. 0
Answer
B. 20 m
It stops at t = 2 s after covering 10 m, then returns 10 m in the next 2 s. Path = 20 m, displacement = 0. (Distance by areas: ½ × 2 × 10 + ½ × 2 × 10.)

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