One-Dimensional Motion Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A particle moves at \( 18 \, \text{m/s} \) and decelerates at \( 4 \, \text{m/s}^2 \) for \( 2 \, \text{s} \), then accelerates at \( 3 \, \text{m/s}^2 \) until its speed is \( 18 \, \text{m/s} \). What is the total distance covered?

Phase 1: \( v = 18 - 4 \cdot 2 = 10 \, \text{m/s} \), \( x_1 = 18 \cdot 2 - \frac{1}{2} \cdot 4 \cdot (2)^2 = 36 - 8 = 28 \, \text{m} \).

Phase 2: \( 18 = 10 + 3 t \Rightarrow t = 2.67 \, \text{s} \), \( x_2 = 10 \cdot 2.67 + \frac{1}{2} \cdot 3 \cdot (2.67)^2 = 26.7 + 10.68 = 37.38 \, \text{m} \).

Total = \( 28 + 37.38 = 65.38 \, \text{m} \).

65 m
65.38 m
66 m
64 m
2

A body starts from rest and accelerates uniformly at \( 6 \, \text{m/s}^2 \) for \( 3 \, \text{s} \). What is the distance covered?

Use \( x = v_0 t + \frac{1}{2} a t^2 \).

Here, \( v_0 = 0 \), \( a = 6 \, \text{m/s}^2 \), \( t = 3 \, \text{s} \).

Substitute: \( x = 0 \cdot 3 + \frac{1}{2} \cdot 6 \cdot (3)^2 = 0 + 3 \cdot 9 = 27 \, \text{m} \).

The distance covered is \( 27 \, \text{m} \).

18 m
27 m
36 m
45 m
2

A train moving at \( 108 \, \text{km/h} \) decelerates at \( 2 \, \text{m/s}^2 \) for \( 10 \, \text{s} \), then at \( 4 \, \text{m/s}^2 \) until it stops. What is the total time taken to stop?

Speed: \( 108 \, \text{km/h} = 30 \, \text{m/s} \).

Phase 1: \( v = 30 - 2 \cdot 10 = 10 \, \text{m/s} \).

Phase 2: \( t = \frac{10}{4} = 2.5 \, \text{s} \).

Total time = \( 10 + 2.5 = 12.5 \, \text{s} \).

12.5 s
13 s
12 s
14 s
1

A man running at \( 6 \, \text{m/s} \) throws a ball horizontally at \( 10 \, \text{m/s} \) relative to himself. If it lands \( 12 \, \text{m} \) ahead of the throw point, what is the time of flight? (Take \( g = 10 \, \text{m/s}^2 \))

Ball’s speed relative to ground = \( 6 + 10 = 16 \, \text{m/s} \).

Relative distance: \( 16 t - 6 t = 12 \Rightarrow 10 t = 12 \Rightarrow t = 1.2 \, \text{s} \).

1 s
1.5 s
1.1 s
1.2 s
4

A stone is dropped from rest. What is its velocity after falling \( 78.4 \, \text{m} \)? (Take \( g = 9.8 \, \text{m/s}^2 \))

Use \( v^2 = v_0^2 + 2 g y \). Here, \( v_0 = 0 \), \( g = 9.8 \, \text{m/s}^2 \), \( y = 78.4 \, \text{m} \).

Substitute: \( v^2 = 0 + 2 \cdot 9.8 \cdot 78.4 = 1536.64 \Rightarrow v = \sqrt{1536.64} \approx 39.2 \, \text{m/s} \).

The velocity is \( 39.2 \, \text{m/s} \).

35 m/s
40 m/s
39.2 m/s
30 m/s
3

A stone is thrown upwards with a speed of \( 14 \, \text{m/s} \). What is its velocity after \( 1.5 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v = v_0 + a t \). Here, \( v_0 = 14 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \), \( t = 1.5 \, \text{s} \).

Substitute: \( v = 14 - 10 \cdot 1.5 = 14 - 15 = -1 \, \text{m/s} \).

The velocity is \( -1 \, \text{m/s} \) (downward).

0 m/s
-1 m/s
1 m/s
-2 m/s
2

A cyclist moving at \( 10 \, \text{m/s} \) accelerates at \( 1 \, \text{m/s}^2 \) for \( 5 \, \text{s} \), then decelerates at \( 2 \, \text{m/s}^2 \) to rest. What is the total distance covered?

Phase 1: \( v = 10 + 1 \cdot 5 = 15 \, \text{m/s} \), \( x_1 = 10 \cdot 5 + \frac{1}{2} \cdot 1 \cdot (5)^2 = 50 + 12.5 = 62.5 \, \text{m} \).

Phase 2: \( t = \frac{15}{2} = 7.5 \, \text{s} \), \( x_2 = 15 \cdot 7.5 - \frac{1}{2} \cdot 2 \cdot (7.5)^2 = 112.5 - 56.25 = 56.25 \, \text{m} \).

Total = \( 62.5 + 56.25 = 118.75 \, \text{m} \).

115 m
118.75 m
120 m
125 m
2

A car moving at \( 27 \, \text{m/s} \) decelerates uniformly to rest in \( 9 \, \text{s} \). What is the distance covered?

Find \( a = \frac{v - v_0}{t} = \frac{0 - 27}{9} = -3 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 27 \cdot 9 + \frac{1}{2} (-3) (9)^2 = 243 - 121.5 = 121.5 \, \text{m} \).

The distance covered is \( 121.5 \, \text{m} \).

100 m
150 m
121.5 m
135 m
3

A car moving at \( 60 \, \text{m/s} \) decelerates uniformly to rest over \( 150 \, \text{m} \). What is the time taken to stop?

Use \( v^2 = v_0^2 + 2 a x \) to find \( a \): \( 0 = (60)^2 + 2 a (150) \Rightarrow 0 = 3600 + 300 a \Rightarrow a = -12 \, \text{m/s}^2 \).

Then, \( v = v_0 + a t \): \( 0 = 60 - 12 t \Rightarrow t = 5 \, \text{s} \).

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

4 s
6 s
7 s
5 s
4

A train accelerates uniformly from rest to \( 15 \, \text{m/s} \) over a distance of \( 37.5 \, \text{m} \). What is its acceleration?

Use \( v^2 = v_0^2 + 2 a x \). Here, \( v_0 = 0 \), \( v = 15 \, \text{m/s} \), \( x = 37.5 \, \text{m} \).

Substitute: \( (15)^2 = 0 + 2 a (37.5) \Rightarrow 225 = 75 a \Rightarrow a = 3 \, \text{m/s}^2 \).

The acceleration is \( 3 \, \text{m/s}^2 \).

2 m/s²
4 m/s²
1.5 m/s²
3 m/s²
4

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