One-Dimensional Motion Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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

Acceleration: \( v = 0 + 2 \cdot 10 = 20 \, \text{m/s} \), \( x_1 = \frac{1}{2} \cdot 2 \cdot (10)^2 = 100 \, \text{m} \).

Deceleration: \( t = \frac{20 - 0}{1} = 20 \, \text{s} \), \( x_2 = 20 \cdot 20 + \frac{1}{2} (-1) (20)^2 = 400 - 200 = 200 \, \text{m} \).

Total distance = \( 100 + 200 = 300 \, \text{m} \).

250 m
300 m
350 m
400 m
2

A car accelerates uniformly from \( 10 \, \text{m/s} \) to \( 30 \, \text{m/s} \) over a distance of \( 80 \, \text{m} \). What is the acceleration?

Use \( v^2 = v_0^2 + 2 a x \).

Here, \( v = 30 \, \text{m/s} \), \( v_0 = 10 \, \text{m/s} \), \( x = 80 \, \text{m} \).

Substitute: \( (30)^2 = (10)^2 + 2 a (80) \Rightarrow 900 = 100 + 160 a \Rightarrow 160 a = 800 \Rightarrow a = 5 \, \text{m/s}^2 \).

The acceleration is \( 5 \, \text{m/s}^2 \).

2.5 m/s²
4 m/s²
6 m/s²
5 m/s²
4

A car accelerates from rest at \( 4 \, \text{m/s}^2 \) for \( 6 \, \text{s} \), then decelerates at \( 3 \, \text{m/s}^2 \) until its speed is \( 12 \, \text{m/s} \). What is the total distance covered?

Phase 1: \( v = 4 \cdot 6 = 24 \, \text{m/s} \), \( x_1 = \frac{1}{2} \cdot 4 \cdot (6)^2 = 72 \, \text{m} \).

Phase 2: \( 12 = 24 - 3 t \Rightarrow t = 4 \, \text{s} \), \( x_2 = 24 \cdot 4 - \frac{1}{2} \cdot 3 \cdot (4)^2 = 96 - 24 = 72 \, \text{m} \).

Total = \( 72 + 72 = 144 \, \text{m} \).

140 m
144 m
150 m
136 m
1

A body moving with an initial velocity of \( 20 \, \text{m/s} \) comes to rest after traveling \( 40 \, \text{m} \) under uniform retardation. What is the magnitude of its acceleration?

Use \( v^2 = v_0^2 + 2 a x \). Here, \( v = 0 \), \( v_0 = 20 \, \text{m/s} \), \( x = 40 \, \text{m} \).

Substitute: \( 0 = (20)^2 + 2 a (40) \Rightarrow 0 = 400 + 80 a \Rightarrow 80 a = -400 \Rightarrow a = -5 \, \text{m/s}^2 \).

Magnitude of acceleration is \( 5 \, \text{m/s}^2 \).

5 m/s²
10 m/s²
2.5 m/s²
7.5 m/s²
1

A ball is thrown upwards with a speed of \( 36 \, \text{m/s} \). 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, \( v = 0 \), \( v_0 = 36 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = (36)^2 + 2 (-10) h \Rightarrow 0 = 1296 - 20 h \Rightarrow h = \frac{1296}{20} = 64.8 \, \text{m} \).

The maximum height is \( 64.8 \, \text{m} \).

60 m
70 m
64.8 m
50 m
3

A cyclist accelerates from \( 10 \, \text{m/s} \) at \( 2 \, \text{m/s}^2 \) for \( 3 \, \text{s} \), then decelerates at \( 1.5 \, \text{m/s}^2 \) until its speed is \( 12 \, \text{m/s} \). What is the total distance covered?

Phase 1: \( v = 10 + 2 \cdot 3 = 16 \, \text{m/s} \), \( x_1 = 10 \cdot 3 + \frac{1}{2} \cdot 2 \cdot (3)^2 = 30 + 9 = 39 \, \text{m} \).

Phase 2: \( 12 = 16 - 1.5 t \Rightarrow t = 2.67 \, \text{s} \), \( x_2 = 16 \cdot 2.67 - \frac{1}{2} \cdot 1.5 \cdot (2.67)^2 = 42.72 - 5.34 = 37.38 \, \text{m} \).

Total = \( 39 + 37.38 = 76.38 \, \text{m} \).

75 m
76.38 m
78 m
74 m
2

A person walks \( 10 \, \text{m} \) in \( 4 \, \text{s} \) and then walks back \( 2 \, \text{m} \) in \( 1 \, \text{s} \). What is the average speed?

Average speed = total distance / total time.

Total distance = \( 10 \, \text{m} + 2 \, \text{m} = 12 \, \text{m} \), total time = \( 4 \, \text{s} + 1 \, \text{s} = 5 \, \text{s} \).

Average speed = \( \frac{12}{5} = 2.4 \, \text{m/s} \).

2 m/s
3 m/s
2.4 m/s
1.8 m/s
3

A train accelerates uniformly from rest to \( 10 \, \text{m/s} \) over a distance of \( 25 \, \text{m} \). What is the acceleration?

Use \( v^2 = v_0^2 + 2 a x \). Here, \( v_0 = 0 \), \( v = 10 \, \text{m/s} \), \( x = 25 \, \text{m} \).

Substitute: \( (10)^2 = 0 + 2 a (25) \Rightarrow 100 = 50 a \Rightarrow a = 2 \, \text{m/s}^2 \).

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

1 m/s²
3 m/s²
4 m/s²
2 m/s²
4

An object falls freely from rest. What is its velocity after traveling \( 4.9 \, \text{m} \)? (Take \( g = 9.8 \, \text{m/s}^2 \))

Use \( v^2 = v_0^2 + 2 g y \). Here, \( v_0 = 0 \), \( g = 9.8 \, \text{m/s}^2 \), \( y = 4.9 \, \text{m} \).

Substitute: \( v^2 = 0 + 2 \cdot 9.8 \cdot 4.9 = 96.04 \Rightarrow v = \sqrt{96.04} \approx 9.8 \, \text{m/s} \).

The velocity is \( 9.8 \, \text{m/s} \).

4.9 m/s
19.6 m/s
9.8 m/s
14.7 m/s
3

Two balls are dropped from a height of \( 180 \, \text{m} \), with a \( 1.5 \, \text{s} \) interval. What is their separation when the first ball hits the ground? (Take \( g = 10 \, \text{m/s}^2 \))

First ball: \( 180 = \frac{1}{2} \cdot 10 \cdot t^2 \Rightarrow t^2 = 36 \Rightarrow t = 6 \, \text{s} \).

Second ball (after 4.5 s): \( y_2 = \frac{1}{2} \cdot 10 \cdot (4.5)^2 = 101.25 \, \text{m} \).

Separation = \( 180 - 101.25 = 78.75 \, \text{m} \).

75 m
80 m
78.75 m
85 m
3

A stone falls from a height of \( 80 \, \text{m} \) on a planet and takes \( 4 \, \text{s} \) to reach the ground. What is the gravitational acceleration on this planet?

\( h = \frac{1}{2} g t^2 \), \( 80 = \frac{1}{2} g (4)^2 \Rightarrow 80 = 8 g \Rightarrow g = 10 \, \text{m/s}^2 \).

9 m/s²
11 m/s²
8 m/s²
10 m/s²
4

A ball is thrown upwards at \( 50 \, \text{m/s} \) from a cliff. It returns to the cliff’s height after \( 12 \, \text{s} \). What is the height of the cliff? (Take \( g = 10 \, \text{m/s}^2 \))

Time to max height: \( t = \frac{50}{10} = 5 \, \text{s} \).

Max height: \( h_{\text{max}} = 50 \cdot 5 - \frac{1}{2} \cdot 10 \cdot (5)^2 = 250 - 125 = 125 \, \text{m} \).

Time down to cliff = \( 12 - 5 = 7 \, \text{s} \). Distance fallen: \( h = \frac{1}{2} \cdot 10 \cdot (7)^2 = 245 \, \text{m} \).

Cliff height = \( 245 - 125 = 120 \, \text{m} \).

100 m
150 m
120 m
130 m
3

A cyclist moves at a constant speed of \( 15 \, \text{km/h} \) for \( 12 \, \text{minutes} \). What is the distance covered?

Convert speed: \( 15 \, \text{km/h} = 15 \cdot \frac{1000}{3600} = \frac{25}{6} \, \text{m/s} \).

Time = \( 12 \, \text{min} = 12 \cdot 60 = 720 \, \text{s} \).

Distance \( x = v t = \frac{25}{6} \cdot 720 = 3000 \, \text{m} = 3 \, \text{km} \).

The distance covered is \( 3 \, \text{km} \).

2.5 km
2 km
3.5 km
3 km
4

A balloon descends at \( 5 \, \text{m/s} \) from \( 150 \, \text{m} \) and releases a stone after \( 3 \, \text{s} \). What is the time taken by the stone to hit the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Height at release: \( h = 150 - 5 \cdot 3 = 135 \, \text{m} \).

Stone’s initial velocity = \( -5 \, \text{m/s} \).

\( -135 = -5 t + 5 t^2 \Rightarrow 5 t^2 - 5 t - 135 = 0 \Rightarrow t^2 - t - 27 = 0 \).

Solve: \( t = \frac{1 \pm \sqrt{1 + 108}}{2} = \frac{1 \pm 10.44}{2} \), \( t = 5.72 \, \text{s} \) (positive root).

5.5 s
5.72 s
6 s
5.2 s
2

A stone is dropped from a height of \( 125 \, \text{m} \). 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, \( v_0 = 0 \), \( g = 10 \, \text{m/s}^2 \), \( y = 125 \, \text{m} \).

Substitute: \( v^2 = 0 + 2 \cdot 10 \cdot 125 = 2500 \Rightarrow v = \sqrt{2500} = 50 \, \text{m/s} \).

The velocity is \( 50 \, \text{m/s} \).

40 m/s
50 m/s
60 m/s
30 m/s
2

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