One-Dimensional Motion Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A runner moving at \( 7 \, \text{m/s} \) throws a ball horizontally at \( 14 \, \text{m/s} \) relative to himself. If it lands \( 21 \, \text{m} \) ahead of the throw point, what is the time of flight? (Take \( g = 10 \, \text{m/s}^2 \))

Ball’s speed relative to ground = \( 7 + 14 = 21 \, \text{m/s} \).

Relative distance: \( 21 t - 7 t = 21 \Rightarrow 14 t = 21 \Rightarrow t = 1.5 \, \text{s} \).

1.2 s
1.8 s
1.4 s
1.5 s
4

A ball is thrown upwards at \( 20 \, \text{m/s} \) from a \( 50 \, \text{m} \) tower. What is its speed when it passes \( 20 \, \text{m} \) below the tower’s top? (Take \( g = 10 \, \text{m/s}^2 \))

Displacement \( y = -20 \, \text{m} \), \( v^2 = (20)^2 + 2 \cdot 10 \cdot 20 = 400 + 400 = 800 \).

\( v = \sqrt{800} \approx 28.28 \, \text{m/s} \).

25 m/s
30 m/s
27 m/s
28.28 m/s
4

A ball is thrown upwards at \( 28 \, \text{m/s} \) from the ground. At the same instant, another ball is dropped from \( 80 \, \text{m} \) above. Where do they meet relative to the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Ball 1: \( y_1 = 28 t - 5 t^2 \).

Ball 2: \( y_2 = 80 - 5 t^2 \).

Meet when \( y_1 = y_2 \): \( 28 t - 5 t^2 = 80 - 5 t^2 \Rightarrow 28 t = 80 \Rightarrow t = \frac{20}{7} \approx 2.86 \, \text{s} \).

Height: \( y_1 = 28 \cdot \frac{20}{7} - 5 \cdot \left(\frac{20}{7}\right)^2 = 80 - 5 \cdot \frac{400}{49} = 80 - 40.82 = 39.18 \, \text{m} \).

39.18 m
40 m
35 m
45 m
1

A particle moves at \( 12 \, \text{m/s} \) and decelerates at \( 3 \, \text{m/s}^2 \) for \( 2 \, \text{s} \), then accelerates at \( 4 \, \text{m/s}^2 \) until it regains \( 12 \, \text{m/s} \). What is the total distance covered?

Phase 1: \( v = 12 - 3 \cdot 2 = 6 \, \text{m/s} \), \( x_1 = 12 \cdot 2 - \frac{1}{2} \cdot 3 \cdot (2)^2 = 24 - 6 = 18 \, \text{m} \).

Phase 2: \( 12 = 6 + 4 t \Rightarrow t = 1.5 \, \text{s} \), \( x_2 = 6 \cdot 1.5 + \frac{1}{2} \cdot 4 \cdot (1.5)^2 = 9 + 4.5 = 13.5 \, \text{m} \).

Total = \( 18 + 13.5 = 31.5 \, \text{m} \).

30 m
31.5 m
35 m
28 m
2

A stone falls from a height and covers \( 68.6 \, \text{m} \) in the last \( 1.5 \, \text{s} \) of its fall. What is the total height? (Take \( g = 9.8 \, \text{m/s}^2 \))

Total time = \( t \), last 1.5 s: \( 68.6 = 9.8 t - 4.9 (t - 1.5)^2 \).

Simplify: \( 68.6 = 9.8 t - 4.9 (t^2 - 3 t + 2.25) \Rightarrow 68.6 = 9.8 t - 4.9 t^2 + 14.7 t - 11.025 \Rightarrow 4.9 t^2 - 24.5 t + 79.625 = 0 \).

Solve: \( t = \frac{24.5 \pm \sqrt{600.25 - 1558.025}}{9.8} \), adjust: \( 68.6 = 4.9 (2 t - 1.5) \Rightarrow 14 = 2 t - 1.5 \Rightarrow t = 7.75 \, \text{s} \).

Height: \( h = \frac{1}{2} \cdot 9.8 \cdot (7.75)^2 = 294.31 \, \text{m} \).

290 m
300 m
294.31 m
280 m
3

A bus starts from rest and moves with a uniform acceleration of \( 1.5 \, \text{m/s}^2 \) for \( 8 \, \text{s} \). What is the final velocity of the bus?

Use the equation \( v = v_0 + a t \).

Here, initial velocity \( v_0 = 0 \), acceleration \( a = 1.5 \, \text{m/s}^2 \), time \( t = 8 \, \text{s} \).

Substitute: \( v = 0 + 1.5 \cdot 8 = 12 \, \text{m/s} \).

The final velocity is \( 12 \, \text{m/s} \).

12 m/s
15 m/s
10 m/s
18 m/s
1

A person walks \( 8 \, \text{m} \) in \( 2 \, \text{s} \) and then walks back \( 4 \, \text{m} \) in \( 1 \, \text{s} \). What is the average speed?

Average speed = total distance / total time.

Total distance = \( 8 \, \text{m} + 4 \, \text{m} = 12 \, \text{m} \), total time = \( 2 \, \text{s} + 1 \, \text{s} = 3 \, \text{s} \).

Average speed = \( \frac{12}{3} = 4 \, \text{m/s} \).

2 m/s
3 m/s
4 m/s
5 m/s
3

A train accelerates uniformly from \( 8 \, \text{m/s} \) to \( 20 \, \text{m/s} \) in \( 6 \, \text{s} \). What is the distance covered?

Find \( a = \frac{v - v_0}{t} = \frac{20 - 8}{6} = 2 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 8 \cdot 6 + \frac{1}{2} \cdot 2 \cdot (6)^2 = 48 + 36 = 84 \, \text{m} \).

The distance covered is \( 84 \, \text{m} \).

72 m
90 m
80 m
84 m
4

A stone falls from a height of \( 98 \, \text{m} \) on a planet and takes \( 3.5 \, \text{s} \) to reach the ground. What is the gravitational acceleration on this planet?

\( h = \frac{1}{2} g t^2 \), \( 98 = \frac{1}{2} g (3.5)^2 \Rightarrow 98 = 6.125 g \Rightarrow g = 16 \, \text{m/s}^2 \).

15 m/s²
17 m/s²
14 m/s²
16 m/s²
4

A truck moving at \( 15 \, \text{m/s} \) decelerates uniformly and stops after \( 30 \, \text{m} \). How long does it take to stop?

Use \( v^2 = v_0^2 + 2 a x \) to find \( a \): \( 0 = (15)^2 + 2 a (30) \Rightarrow 0 = 225 + 60 a \Rightarrow a = -3.75 \, \text{m/s}^2 \).

Then, \( v = v_0 + a t \): \( 0 = 15 - 3.75 t \Rightarrow t = \frac{15}{3.75} = 4 \, \text{s} \).

The time taken is \( 4 \, \text{s} \).

2 s
6 s
4 s
8 s
3

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