One-Dimensional Motion Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Two stones are dropped from a height of \( 245 \, \text{m} \), with a \( 3 \, \text{s} \) interval. What is their separation when the second stone has fallen for \( 4 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

First stone (after 7 s): \( y_1 = \frac{1}{2} \cdot 10 \cdot (7)^2 = 245 \, \text{m} \) (hits ground).

Second stone (after 4 s): \( y_2 = \frac{1}{2} \cdot 10 \cdot (4)^2 = 80 \, \text{m} \).

Separation = \( 245 - 80 = 165 \, \text{m} \).

160 m
170 m
165 m
175 m
3

A ball is thrown upwards at \( 18 \, \text{m/s} \) from a \( 32 \, \text{m} \) tower. What is its speed when it passes the tower’s base on the way down? (Take \( g = 10 \, \text{m/s}^2 \))

\( v^2 = v_0^2 + 2 g h \), \( v^2 = (18)^2 + 2 \cdot 10 \cdot 32 = 324 + 640 = 964 \).

\( v = \sqrt{964} \approx 31.05 \, \text{m/s} \).

30 m/s
32 m/s
28 m/s
31.05 m/s
4

A particle moves with a uniform acceleration of \( 2 \, \text{m/s}^2 \) starting from rest. What is its velocity after traveling \( 18 \, \text{m} \)?

Use \( v^2 = v_0^2 + 2 a x \). Here, \( v_0 = 0 \), \( a = 2 \, \text{m/s}^2 \), \( x = 18 \, \text{m} \).

Substitute: \( v^2 = 0 + 2 \cdot 2 \cdot 18 = 72 \Rightarrow v = \sqrt{72} = 6\sqrt{2} \approx 8.48 \, \text{m/s} \).

Approximate value is \( 8.5 \, \text{m/s} \).

6 m/s
8.5 m/s
10 m/s
12 m/s
2

An object is dropped from a height of \( 19.6 \, \text{m} \) above the ground. How long does it take to reach the ground? (Take \( g = 9.8 \, \text{m/s}^2 \))

For free fall, use \( y = \frac{1}{2} g t^2 \). Here, \( y = 19.6 \, \text{m} \), \( g = 9.8 \, \text{m/s}^2 \).

Substitute: \( 19.6 = \frac{1}{2} \cdot 9.8 \cdot t^2 \Rightarrow 19.6 = 4.9 t^2 \Rightarrow t^2 = 4 \Rightarrow t = 2 \, \text{s} \).

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

2 s
1 s
3 s
4 s
1

A ball is thrown upwards with a speed of \( 40 \, \text{m/s} \). What is the maximum height it reaches? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v^2 = v_0^2 + 2 a h \). At max height, \( v = 0 \), \( v_0 = 40 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = (40)^2 + 2 (-10) h \Rightarrow 0 = 1600 - 20 h \Rightarrow h = \frac{1600}{20} = 80 \, \text{m} \).

The maximum height is \( 80 \, \text{m} \).

40 m
60 m
80 m
100 m
3

A particle accelerates uniformly from rest to \( 18 \, \text{m/s} \) in \( 9 \, \text{s} \). What is the distance covered?

Find \( a = \frac{v - v_0}{t} = \frac{18 - 0}{9} = 2 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 0 \cdot 9 + \frac{1}{2} \cdot 2 \cdot (9)^2 = 0 + 1 \cdot 81 = 81 \, \text{m} \).

The distance covered is \( 81 \, \text{m} \).

72 m
81 m
90 m
54 m
2

A ball is thrown vertically upwards with a speed of \( 12 \, \text{m/s} \). What is the maximum height reached? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v^2 = v_0^2 + 2 a h \). At max height, \( v = 0 \), \( v_0 = 12 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = (12)^2 + 2 (-10) h \Rightarrow 0 = 144 - 20 h \Rightarrow h = \frac{144}{20} = 7.2 \, \text{m} \).

The maximum height is \( 7.2 \, \text{m} \).

7.2 m
14.4 m
3.6 m
10 m
1

A stone is dropped from a cliff and takes \( 5 \, \text{s} \) to hit the ground. What is the height of the cliff? (Take \( g = 10 \, \text{m/s}^2 \))

For free fall, distance is given by \( y = \frac{1}{2} g t^2 \).

Here, \( g = 10 \, \text{m/s}^2 \), \( t = 5 \, \text{s} \).

Substitute: \( y = \frac{1}{2} \cdot 10 \cdot (5)^2 = 5 \cdot 25 = 125 \, \text{m} \).

The height of the cliff is \( 125 \, \text{m} \).

125 m
100 m
150 m
50 m
1

A stone falls freely from a height and reaches the ground with a velocity of \( 19.6 \, \text{m/s} \). What is the height? (Take \( g = 9.8 \, \text{m/s}^2 \))

Use \( v^2 = v_0^2 + 2 g y \). Here, \( v_0 = 0 \), \( v = 19.6 \, \text{m/s} \), \( g = 9.8 \, \text{m/s}^2 \).

Substitute: \( (19.6)^2 = 0 + 2 \cdot 9.8 \cdot y \Rightarrow 384.16 = 19.6 y \Rightarrow y = 19.6 \, \text{m} \).

The height is \( 19.6 \, \text{m} \).

9.8 m
39.2 m
19.6 m
29.4 m
3

A car moves at \( 18 \, \text{m/s} \) and decelerates at \( 2 \, \text{m/s}^2 \) for \( 4 \, \text{s} \), then accelerates at \( 3 \, \text{m/s}^2 \) for \( 5 \, \text{s} \). What is the net displacement?

Phase 1: \( v = 18 - 2 \cdot 4 = 10 \, \text{m/s} \), \( x_1 = 18 \cdot 4 - \frac{1}{2} \cdot 2 \cdot (4)^2 = 72 - 16 = 56 \, \text{m} \).

Phase 2: \( v = 10 + 3 \cdot 5 = 25 \, \text{m/s} \), \( x_2 = 10 \cdot 5 + \frac{1}{2} \cdot 3 \cdot (5)^2 = 50 + 37.5 = 87.5 \, \text{m} \).

Total = \( 56 + 87.5 = 143.5 \, \text{m} \).

140 m
150 m
145 m
143.5 m
4

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