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  4. Motion In A Straight Line Chapter-Wise Test 2

Motion In A Straight Line Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A car accelerates from rest at \( 2.5 \, \text{m/s}^2 \) until it reaches \( 25 \, \text{m/s} \), then immediately decelerates at \( 5 \, \text{m/s}^2 \) to rest. What is the total time taken?

Acceleration: \( t_1 = \frac{25}{2.5} = 10 \, \text{s} \).

Deceleration: \( t_2 = \frac{25}{5} = 5 \, \text{s} \).

Total time = \( 10 + 5 = 15 \, \text{s} \).

12 s
15 s
18 s
14 s
1

An elevator ascends at \( 3 \, \text{m/s} \) for \( 5 \, \text{s} \), then accelerates at \( 2 \, \text{m/s}^2 \) for \( 3 \, \text{s} \). A ball is dropped at the end of this period. What is the time taken for the ball to hit the ground if the initial height was \( 20 \, \text{m} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Phase 1: \( h_1 = 3 \cdot 5 = 15 \, \text{m} \).

Phase 2: \( v = 3 + 2 \cdot 3 = 9 \, \text{m/s} \), \( h_2 = 3 \cdot 3 + \frac{1}{2} \cdot 2 \cdot (3)^2 = 9 + 9 = 18 \, \text{m} \).

Total height at drop = \( 20 + 15 + 18 = 53 \, \text{m} \), initial velocity of ball = \( 9 \, \text{m/s} \) upward.

\( -53 = 9 t - 5 t^2 \Rightarrow 5 t^2 - 9 t - 53 = 0 \).

Solve: \( t = \frac{9 \pm \sqrt{81 + 1060}}{10} = \frac{9 \pm 33.79}{10} \), \( t = 4.28 \, \text{s} \) (positive root).

4.28 s
4.5 s
4 s
5 s
1

A rocket accelerates from rest at \( 4 \, \text{m/s}^2 \) for \( 6 \, \text{s} \), then moves at constant speed for \( 5 \, \text{s} \). What is the total distance covered?

Phase 1: \( v = 4 \cdot 6 = 24 \, \text{m/s} \), \( x_1 = \frac{1}{2} \cdot 4 \cdot (6)^2 = 72 \, \text{m} \).

Phase 2: \( x_2 = 24 \cdot 5 = 120 \, \text{m} \).

Total = \( 72 + 120 = 192 \, \text{m} \).

180 m
200 m
190 m
192 m
4

A ball is thrown vertically upwards with an initial velocity of \( 10 \, \text{m/s} \). What is the total time of flight? (Take \( g = 10 \, \text{m/s}^2 \))

Time to reach max height: \( v = v_0 + a t \), \( 0 = 10 - 10 t \Rightarrow t = 1 \, \text{s} \).

Total time of flight = time up + time down = \( 1 \, \text{s} + 1 \, \text{s} = 2 \, \text{s} \).

The total time is \( 2 \, \text{s} \).

2 s
1 s
3 s
4 s
1

A man walking at \( 4 \, \text{m/s} \) throws a ball forward at \( 12 \, \text{m/s} \) relative to himself. If the ball lands \( 10 \, \text{m} \) ahead of the throw point, how long was it in the air? (Take \( g = 10 \, \text{m/s}^2 \))

Ball’s speed relative to ground = \( 12 + 4 = 16 \, \text{m/s} \).

Man moves \( 4 t \), ball moves \( 16 t \), distance ahead = \( 16 t - 4 t = 10 \Rightarrow 12 t = 10 \Rightarrow t = \frac{5}{6} \, \text{s} \).

0.5 s
1 s
0.75 s
0.833 s
4

A body accelerates uniformly from \( 8 \, \text{m/s} \) to \( 16 \, \text{m/s} \) in \( 4 \, \text{s} \). What is the distance traveled?

Find \( a = \frac{v - v_0}{t} = \frac{16 - 8}{4} = 2 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 8 \cdot 4 + \frac{1}{2} \cdot 2 \cdot (4)^2 = 32 + 16 = 48 \, \text{m} \).

The distance traveled is \( 48 \, \text{m} \).

32 m
48 m
40 m
56 m
2

A ball is thrown upwards at \( 45 \, \text{m/s} \) from a \( 30 \, \text{m} \) tower. What is the time taken to reach a point \( 15 \, \text{m} \) above the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Displacement \( y = -15 \, \text{m} \), \( -15 = 45 t - 5 t^2 \Rightarrow 5 t^2 - 45 t - 15 = 0 \Rightarrow t^2 - 9 t - 3 = 0 \).

Solve: \( t = \frac{9 \pm \sqrt{81 + 12}}{2} = \frac{9 \pm 9.64}{2} \), \( t = 9.32 \, \text{s} \) (downward path).

9 s
9.5 s
9.32 s
8.5 s
3

A drone ascends vertically at \( 5 \, \text{m/s} \) for \( 8 \, \text{s} \), then releases a payload that hits the ground \( 6 \, \text{s} \) later. What was the initial height of the drone above the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Height after ascent: \( h_1 = 5 \cdot 8 = 40 \, \text{m} \).

Payload’s initial velocity = \( 536 \, \text{m/s} \) upward.

Total height \( h = h_0 + 40 \), use \( y = v_0 t - \frac{1}{2} g t^2 \), \( y = -h \).

\( -h = 5 \cdot 6 - 5 \cdot (6)^2 \Rightarrow -h = 30 - 180 = -150 \Rightarrow h = 150 \, \text{m} \).

Initial height \( h_0 = 150 - 40 = 110 \, \text{m} \).

110 m
100 m
120 m
130 m
1

A rocket descends at \( 8 \, \text{m/s} \) from a height of \( 120 \, \text{m} \) and releases a payload \( 3 \, \text{s} \) later. How long does the payload take to hit the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Height at release: \( h = 120 - 8 \cdot 3 = 96 \, \text{m} \).

Payload’s initial velocity = \( -8 \, \text{m/s} \).

\( -96 = -8 t + 5 t^2 \Rightarrow 5 t^2 - 8 t - 96 = 0 \).

Solve: \( t = \frac{8 \pm \sqrt{64 + 1920}}{10} = \frac{8 \pm 44.5}{10} \), \( t = 5.25 \, \text{s} \) (positive root).

5 s
5.25 s
5.5 s
4.8 s
2

A car accelerates from rest at \( 3 \, \text{m/s}^2 \) for \( 4 \, \text{s} \), then moves at constant speed for \( 6 \, \text{s} \), and decelerates to rest at \( 2 \, \text{m/s}^2 \). What is the total distance?

Phase 1: \( v = 3 \cdot 4 = 12 \, \text{m/s} \), \( x_1 = \frac{1}{2} \cdot 3 \cdot (4)^2 = 24 \, \text{m} \).

Phase 2: \( x_2 = 12 \cdot 6 = 72 \, \text{m} \).

Phase 3: \( t = \frac{12}{2} = 6 \, \text{s} \), \( x_3 = 12 \cdot 6 - \frac{1}{2} \cdot 2 \cdot (6)^2 = 72 - 36 = 36 \, \text{m} \).

Total distance = \( 24 + 72 + 36 = 132 \, \text{m} \).

120 m
140 m
132 m
150 m
3

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