One-Dimensional Motion Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A lift ascends at \( 4 \, \text{m/s} \) and decelerates at \( 2 \, \text{m/s}^2 \) to stop. If a ball is dropped \( 1.5 \, \text{s} \) after deceleration starts, how far does it fall relative to the ground in \( 1 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Lift stops in \( \frac{4}{2} = 2 \, \text{s} \).

At \( t = 1.5 \, \text{s} \): \( v_{\text{lift}} = 4 - 2 \cdot 1.5 = 1 \, \text{m/s} \), \( y_{\text{lift}} = 4 \cdot 1.5 - \frac{1}{2} \cdot 2 \cdot (1.5)^2 = 6 - 2.25 = 3.75 \, \text{m} \).

Ball: \( y_{\text{ball}} = 1 \cdot 1 + \frac{1}{2} \cdot 10 \cdot (1)^2 = 1 + 5 = 6 \, \text{m} \) from drop point.

Lift at \( t = 2.5 \, \text{s} \): \( y = 4 \cdot 2 - \frac{1}{2} \cdot 2 \cdot (2)^2 = 8 - 4 = 4 \, \text{m} \), then constant.

Ball’s position = \( 3.75 + 6 = 9.75 \, \text{m} \), distance = \( 9.75 - 4 = 5.75 \, \text{m} \).

5 m
5.75 m
6 m
4.5 m
2

A body accelerates uniformly from rest to \( 14 \, \text{m/s} \) over a distance of \( 49 \, \text{m} \). What is the acceleration?

Use \( v^2 = v_0^2 + 2 a x \). Here, \( v_0 = 0 \), \( v = 14 \, \text{m/s} \), \( x = 49 \, \text{m} \).

Substitute: \( (14)^2 = 0 + 2 a (49) \Rightarrow 196 = 98 a \Rightarrow a = 2 \, \text{m/s}^2 \).

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

1 m/s²
3 m/s²
2 m/s²
4 m/s²
3

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

Find \( a = \frac{v - v_0}{t} = \frac{16 - 0}{8} = 2 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 0 \cdot 8 + \frac{1}{2} \cdot 2 \cdot (8)^2 = 0 + 1 \cdot 64 = 64 \, \text{m} \).

The distance covered is \( 64 \, \text{m} \).

48 m
64 m
80 m
32 m
2

A ball is thrown upwards with a speed of \( 45 \, \text{m/s} \). What is the time to reach the maximum height? (Take \( g = 10 \, \text{m/s}^2 \))

At max height, \( v = 0 \). Use \( v = v_0 + a t \).

Here, \( v_0 = 45 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = 45 - 10 t \Rightarrow 10 t = 45 \Rightarrow t = 4.5 \, \text{s} \).

The time taken is \( 4.5 \, \text{s} \).

3 s
4 s
5 s
4.5 s
4

A ball is dropped from a height of \( 31.25 \, \text{m} \). What is its velocity just before hitting the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v^2 = v_0^2 + 2 g y \). Here, \( v_0 = 0 \), \( g = 10 \, \text{m/s}^2 \), \( y = 31.25 \, \text{m} \).

Substitute: \( v^2 = 0 + 2 \cdot 10 \cdot 31.25 = 625 \Rightarrow v = \sqrt{625} = 25 \, \text{m/s} \).

The velocity is \( 25 \, \text{m/s} \).

20 m/s
25 m/s
30 m/s
15 m/s
2

A ball is thrown upwards with a speed of \( 22 \, \text{m/s} \). What is its velocity after \( 1 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v = v_0 + a t \). Here, \( v_0 = 22 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \), \( t = 1 \, \text{s} \).

Substitute: \( v = 22 - 10 \cdot 1 = 12 \, \text{m/s} \).

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

10 m/s
12 m/s
15 m/s
8 m/s
2

A balloon descends at \( 8 \, \text{m/s} \) and releases a stone. If the stone hits the ground after \( 5 \, \text{s} \), what was the height of the balloon? (Take \( g = 10 \, \text{m/s}^2 \))

Initial velocity = \( -8 \, \text{m/s} \) (upward positive).

\( y = v_0 t + \frac{1}{2} g t^2 \), \( y = -8 \cdot 5 + \frac{1}{2} \cdot 10 \cdot (5)^2 = -40 + 125 = 85 \, \text{m} \).

Height = \( 85 \, \text{m} \).

80 m
85 m
90 m
75 m
2

A stone is dropped from a balloon ascending at \( 12 \, \text{m/s} \). If it hits the ground after \( 6 \, \text{s} \), what was the height of the balloon at the moment of release? (Take \( g = 10 \, \text{m/s}^2 \))

Initial velocity of stone = \( 12 \, \text{m/s} \) upward. \( y = v_0 t - \frac{1}{2} g t^2 \).

Ground level \( y = -h \): \( -h = 12 \cdot 6 - \frac{1}{2} \cdot 10 \cdot (6)^2 = 72 - 180 = -108 \).

Height \( h = 108 \, \text{m} \).

100 m
120 m
90 m
108 m
4

A ball is thrown upwards with a speed of \( 55 \, \text{m/s} \). What is the time to reach the maximum height? (Take \( g = 10 \, \text{m/s}^2 \))

At max height, \( v = 0 \). Use \( v = v_0 + a t \).

Here, \( v_0 = 55 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = 55 - 10 t \Rightarrow 10 t = 55 \Rightarrow t = 5.5 \, \text{s} \).

The time taken is \( 5.5 \, \text{s} \).

5 s
6 s
5.5 s
4.5 s
3

A lift descends at \( 3 \, \text{m/s} \) and decelerates at \( 2 \, \text{m/s}^2 \) to stop. If a ball is dropped \( 1 \, \text{s} \) after deceleration begins, how far does it fall relative to the ground in \( 2 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

At drop: \( v_{\text{lift}} = 3 - 2 \cdot 1 = 1 \, \text{m/s} \) downward.

Ball: \( y_{\text{ball}} = 1 \cdot 2 + \frac{1}{2} \cdot 10 \cdot (2)^2 = 2 + 20 = 22 \, \text{m} \).

Lift stops in \( \frac{3}{2} = 1.5 \, \text{s} \), distance = \( 3 \cdot 1.5 - \frac{1}{2} \cdot 2 \cdot (1.5)^2 = 4.5 - 2.25 = 2.25 \, \text{m} \), then stationary.

After 2 s, lift at \( 2.25 \, \text{m} \), ball at \( 22 \, \text{m} \), distance = \( 22 - 2.25 = 19.75 \, \text{m} \).

20 m
19.75 m
22 m
18 m
2

A ball is dropped from a height of \( 100 \, \text{m} \) on a planet where \( g = 5 \, \text{m/s}^2 \). What is the ratio of the distance covered in the 2nd second to the 1st second?

Distance in \( n \)-th second: \( d_n = g \cdot \frac{2n - 1}{2} \).

1st second: \( d_1 = 5 \cdot \frac{2 \cdot 1 - 1}{2} = 2.5 \, \text{m} \).

2nd second: \( d_2 = 5 \cdot \frac{2 \cdot 2 - 1}{2} = 7.5 \, \text{m} \).

Ratio: \( \frac{d_2}{d_1} = \frac{7.5}{2.5} = 3 \).

2
3
1.5
4
2

A ball is dropped from a tower. After \( 2 \, \text{s} \), its velocity is \( 19.6 \, \text{m/s} \). What is the acceleration due to gravity?

Use \( v = v_0 + a t \). Here, \( v_0 = 0 \), \( v = 19.6 \, \text{m/s} \), \( t = 2 \, \text{s} \).

Substitute: \( 19.6 = 0 + a \cdot 2 \Rightarrow a = \frac{19.6}{2} = 9.8 \, \text{m/s}^2 \).

The acceleration is \( 9.8 \, \text{m/s}^2 \).

4.9 m/s²
9.8 m/s²
19.6 m/s²
10 m/s²
2

A body accelerates uniformly from \( 5 \, \text{m/s} \) to \( 15 \, \text{m/s} \) in \( 5 \, \text{s} \). What is the distance covered?

Find \( a = \frac{v - v_0}{t} = \frac{15 - 5}{5} = 2 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 5 \cdot 5 + \frac{1}{2} \cdot 2 \cdot (5)^2 = 25 + 25 = 50 \, \text{m} \).

The distance covered is \( 50 \, \text{m} \).

25 m
50 m
75 m
40 m
2

A stone is dropped from rest. What is the distance covered in the 4th second of its fall? (Take \( g = 10 \, \text{m/s}^2 \))

Distance in \( n \)-th second = \( g \cdot \frac{2n - 1}{2} \). For 4th second, \( n = 4 \).

Substitute: \( d = 10 \cdot \frac{2 \cdot 4 - 1}{2} = 10 \cdot \frac{7}{2} = 35 \, \text{m} \).

The distance covered is \( 35 \, \text{m} \).

25 m
30 m
40 m
35 m
4

A balloon ascends at \( 7 \, \text{m/s} \) and releases a stone after \( 4 \, \text{s} \). If the stone hits the ground \( 5 \, \text{s} \) later, what was the initial height of the balloon? (Take \( g = 10 \, \text{m/s}^2 \))

Height at release: \( h_1 = 7 \cdot 4 = 28 \, \text{m} \).

Stone: \( h_{\text{total}} = h_0 + 28 \), \( -h_{\text{total}} = 7 \cdot 5 - \frac{1}{2} \cdot 10 \cdot (5)^2 = 35 - 125 = -90 \).

\( h_{\text{total}} = 90 \, \text{m} \), \( h_0 = 90 - 28 = 62 \, \text{m} \).

60 m
62 m
65 m
58 m
2

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