One-Dimensional Motion Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A runner moves at a constant speed of \( 8 \, \text{m/s} \) for \( 25 \, \text{s} \). What is the distance covered by the runner?

For constant speed, distance \( x = v t \).

Here, \( v = 8 \, \text{m/s} \), \( t = 25 \, \text{s} \).

Substitute: \( x = 8 \cdot 25 = 200 \, \text{m} \).

The distance covered is \( 200 \, \text{m} \).

200 m
150 m
250 m
100 m
1

A car moving at \( 25 \, \text{m/s} \) decelerates uniformly to rest over \( 62.5 \, \text{m} \). What is the time taken to stop?

Use \( v^2 = v_0^2 + 2 a x \) to find \( a \): \( 0 = (25)^2 + 2 a (62.5) \Rightarrow 0 = 625 + 125 a \Rightarrow a = -5 \, \text{m/s}^2 \).

Then, \( v = v_0 + a t \): \( 0 = 25 - 5 t \Rightarrow t = 5 \, \text{s} \).

The time taken is \( 5 \, \text{s} \).

4 s
6 s
3 s
5 s
4

A particle moves at \( 15 \, \text{m/s} \) and decelerates at \( 3 \, \text{m/s}^2 \) for \( 3 \, \text{s} \), then accelerates at \( 2 \, \text{m/s}^2 \) until its speed is \( 15 \, \text{m/s} \). What is the total distance covered?

Phase 1: \( v = 15 - 3 \cdot 3 = 6 \, \text{m/s} \), \( x_1 = 15 \cdot 3 - \frac{1}{2} \cdot 3 \cdot (3)^2 = 45 - 13.5 = 31.5 \, \text{m} \).

Phase 2: \( 15 = 6 + 2 t \Rightarrow t = 4.5 \, \text{s} \), \( x_2 = 6 \cdot 4.5 + \frac{1}{2} \cdot 2 \cdot (4.5)^2 = 27 + 20.25 = 47.25 \, \text{m} \).

Total = \( 31.5 + 47.25 = 78.75 \, \text{m} \).

75 m
78.75 m
80 m
76 m
2

A body starts from rest and accelerates at \( 3 \, \text{m/s}^2 \) for \( 4 \, \text{s} \). What is its final velocity?

Use \( v = v_0 + a t \). Here, \( v_0 = 0 \), \( a = 3 \, \text{m/s}^2 \), \( t = 4 \, \text{s} \).

Substitute: \( v = 0 + 3 \cdot 4 = 12 \, \text{m/s} \).

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

6 m/s
9 m/s
15 m/s
12 m/s
4

A stone is thrown upwards with a speed of \( 18 \, \text{m/s} \). What is its velocity after \( 2 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v = v_0 + a t \). Here, \( v_0 = 18 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \), \( t = 2 \, \text{s} \).

Substitute: \( v = 18 - 10 \cdot 2 = 18 - 20 = -2 \, \text{m/s} \).

The velocity is \( -2 \, \text{m/s} \) (downward).

0 m/s
-2 m/s
2 m/s
-4 m/s
2

A train accelerates uniformly from \( 10 \, \text{m/s} \) to \( 30 \, \text{m/s} \) in \( 5 \, \text{s} \). What is its acceleration?

Use \( a = \frac{v - v_0}{t} \). Here, \( v = 30 \, \text{m/s} \), \( v_0 = 10 \, \text{m/s} \), \( t = 5 \, \text{s} \).

Substitute: \( a = \frac{30 - 10}{5} = \frac{20}{5} = 4 \, \text{m/s}^2 \).

The acceleration is \( 4 \, \text{m/s}^2 \).

4 m/s²
6 m/s²
2 m/s²
8 m/s²
1

A stone is thrown upwards from a \( 60 \, \text{m} \) cliff at \( 20 \, \text{m/s} \). What is the time taken to reach a point \( 20 \, \text{m} \) below the cliff’s edge? (Take \( g = 10 \, \text{m/s}^2 \))

Displacement \( y = -80 \, \text{m} \) (downward positive).

Use \( y = v_0 t + \frac{1}{2} g t^2 \), \( v_0 = -20 \, \text{m/s} \).

\( 80 = -20 t + 5 t^2 \Rightarrow 5 t^2 - 20 t - 80 = 0 \Rightarrow t^2 - 4 t - 16 = 0 \).

Solve: \( t = \frac{4 \pm \sqrt{16 + 64}}{2} = \frac{4 \pm 8.94}{2} \), \( t = 6.47 \, \text{s} \) (positive root).

6.47 s
6 s
7 s
5.5 s
1

A ball is dropped from a height of \( 44.1 \, \text{m} \). What is its velocity just before hitting the ground? (Take \( g = 9.8 \, \text{m/s}^2 \))

Use \( v^2 = v_0^2 + 2 g y \). Here, \( v_0 = 0 \), \( g = 9.8 \, \text{m/s}^2 \), \( y = 44.1 \, \text{m} \).

Substitute: \( v^2 = 0 + 2 \cdot 9.8 \cdot 44.1 = 864.36 \Rightarrow v = \sqrt{864.36} \approx 29.4 \, \text{m/s} \).

The velocity is \( 29.4 \, \text{m/s} \).

19.6 m/s
29.4 m/s
14.7 m/s
39.2 m/s
2

A stone is dropped from a height \( h \) and falls \( 44.1 \, \text{m} \) in the last \( 1 \, \text{s} \) before hitting the ground. What is the height \( h \)? (Take \( g = 9.8 \, \text{m/s}^2 \))

Distance in last second: \( d_n = g \cdot \frac{2n - 1}{2} \), \( 44.1 = 9.8 \cdot \frac{2n - 1}{2} \Rightarrow 2n - 1 = 9 \Rightarrow n = 5 \, \text{s} \).

Total height: \( h = \frac{1}{2} g t^2 = \frac{1}{2} \cdot 9.8 \cdot (5)^2 = 122.5 \, \text{m} \).

120 m
130 m
122.5 m
115 m
3

A car accelerates uniformly from rest at \( 3 \, \text{m/s}^2 \) for \( 7 \, \text{s} \). What is the final velocity?

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

Here, initial velocity \( v_0 = 0 \), acceleration \( a = 3 \, \text{m/s}^2 \), time \( t = 7 \, \text{s} \).

Substitute: \( v = 0 + 3 \cdot 7 = 21 \, \text{m/s} \).

The final velocity is \( 21 \, \text{m/s} \).

21 m/s
18 m/s
24 m/s
15 m/s
1

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