One-Dimensional Motion Chapter-Wise Test 17

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A truck moving at \( 90 \, \text{km/h} \) applies brakes and stops after \( 12 \, \text{s} \). If the deceleration triples after \( 8 \, \text{s} \), what is the total distance covered?

Speed: \( 90 \, \text{km/h} = 25 \, \text{m/s} \).

First 8 s: Let \( a_1 = a \), \( v = 25 - a \cdot 8 \), stops in 12 s.

Last 4 s: \( a_2 = 3a \), \( 0 = (25 - 8a) - 3a \cdot 4 \Rightarrow 25 - 8a = 12a \Rightarrow 25 = 20a \Rightarrow a = 1.25 \, \text{m/s}^2 \).

Distance: \( x_1 = 25 \cdot 8 - \frac{1}{2} \cdot 1.25 \cdot (8)^2 = 200 - 40 = 160 \, \text{m} \).

\( x_2 = (25 - 10) \cdot 4 - \frac{1}{2} \cdot 3.75 \cdot (4)^2 = 60 - 30 = 30 \, \text{m} \).

Total = \( 160 + 30 = 190 \, \text{m} \).

190 m
180 m
200 m
210 m
1

A stone is dropped from rest. What is its velocity after falling \( 24.5 \, \text{m} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v^2 = v_0^2 + 2 g y \). Here, \( v_0 = 0 \), \( g = 10 \, \text{m/s}^2 \), \( y = 24.5 \, \text{m} \).

Substitute: \( v^2 = 0 + 2 \cdot 10 \cdot 24.5 = 490 \Rightarrow v = \sqrt{490} \approx 22.14 \, \text{m/s} \).

Rounded to one decimal, the velocity is \( 22.1 \, \text{m/s} \).

20 m/s
25 m/s
22.1 m/s
15 m/s
3

A car moving at \( 12 \, \text{m/s} \) decelerates uniformly to rest in \( 3 \, \text{s} \). What is the distance covered?

Find \( a = \frac{v - v_0}{t} = \frac{0 - 12}{3} = -4 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 12 \cdot 3 + \frac{1}{2} (-4) (3)^2 = 36 - 18 = 18 \, \text{m} \).

The distance covered is \( 18 \, \text{m} \).

12 m
24 m
18 m
30 m
3

An object falls freely from a height. After \( 3 \, \text{s} \), what is the distance covered? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( y = \frac{1}{2} g t^2 \). Here, \( g = 10 \, \text{m/s}^2 \), \( t = 3 \, \text{s} \).

Substitute: \( y = \frac{1}{2} \cdot 10 \cdot (3)^2 = 5 \cdot 9 = 45 \, \text{m} \).

The distance covered is \( 45 \, \text{m} \).

30 m
15 m
60 m
45 m
4

Two balls are dropped from a height of \( 125 \, \text{m} \), \( 2 \, \text{s} \) apart. How long after the second ball is released do they meet if the first ball rebounds at \( 10 \, \text{m/s} \) upon hitting the ground? (Take \( g = 10 \, \text{m/s}^2 \))

First ball: \( 125 = \frac{1}{2} \cdot 10 \cdot t^2 \Rightarrow t = 5 \, \text{s} \).

Second ball released at \( t = 3 \, \text{s} \), meets after \( t' \) s from release.

First ball after rebound: \( y_1 = 10 (t' + 2) \) (upward), second ball: \( y_2 = \frac{1}{2} \cdot 10 \cdot t'^2 \).

Height relation: \( 125 - y_1 = y_2 \Rightarrow 125 - 10 (t' + 2) = 5 t'^2 \Rightarrow 125 - 10 t' - 20 = 5 t'^2 \Rightarrow 5 t'^2 + 10 t' - 105 = 0 \).

Solve: \( t' = \frac{-10 \pm \sqrt{100 + 2100}}{10} = \frac{-10 \pm 47}{10} \), \( t' = 3.7 \, \text{s} \) (positive root).

3.7 s
4 s
3.5 s
3 s
1

A stone falls freely from a height and reaches the ground with a velocity of \( 34.3 \, \text{m/s} \). What is the height? (Take \( g = 9.8 \, \text{m/s}^2 \))

Use \( v^2 = v_0^2 + 2 g y \). Here, \( v_0 = 0 \), \( v = 34.3 \, \text{m/s} \), \( g = 9.8 \, \text{m/s}^2 \).

Substitute: \( (34.3)^2 = 0 + 2 \cdot 9.8 \cdot y \Rightarrow 1176.49 = 19.6 y \Rightarrow y = \frac{1176.49}{19.6} \approx 60 \, \text{m} \).

The height is \( 60 \, \text{m} \).

50 m
70 m
60 m
40 m
3

A ball is thrown upwards with a speed of \( 44 \, \text{m/s} \). What is the total time of flight? (Take \( g = 10 \, \text{m/s}^2 \))

Time to max height: \( v = v_0 + a t \), \( 0 = 44 - 10 t \Rightarrow t = 4.4 \, \text{s} \).

Total time = time up + time down = \( 4.4 \, \text{s} + 4.4 \, \text{s} = 8.8 \, \text{s} \).

The total time is \( 8.8 \, \text{s} \).

8 s
9 s
8.8 s
10 s
3

A person walks \( 12 \, \text{m} \) in \( 3 \, \text{s} \) and then walks back \( 6 \, \text{m} \) in \( 2 \, \text{s} \). What is the average speed?

Average speed = total distance / total time.

Total distance = \( 12 \, \text{m} + 6 \, \text{m} = 18 \, \text{m} \), total time = \( 3 \, \text{s} + 2 \, \text{s} = 5 \, \text{s} \).

Average speed = \( \frac{18}{5} = 3.6 \, \text{m/s} \).

3 m/s
4 m/s
3.6 m/s
2.4 m/s
3

A car accelerates from \( 10 \, \text{m/s} \) at \( 1.5 \, \text{m/s}^2 \) for \( 6 \, \text{s} \), then moves at constant speed for \( 4 \, \text{s} \). What is the average speed over the entire motion?

Phase 1: \( v = 10 + 1.5 \cdot 6 = 19 \, \text{m/s} \), \( x_1 = 10 \cdot 6 + \frac{1}{2} \cdot 1.5 \cdot (6)^2 = 60 + 27 = 87 \, \text{m} \).

Phase 2: \( x_2 = 19 \cdot 4 = 76 \, \text{m} \).

Total distance = \( 87 + 76 = 163 \, \text{m} \), time = \( 6 + 4 = 10 \, \text{s} \).

Average speed = \( \frac{163}{10} = 16.3 \, \text{m/s} \).

16 m/s
17 m/s
16.3 m/s
15.5 m/s
3

A ball is thrown upwards with a speed of \( 26 \, \text{m/s} \). What is its velocity after \( 2 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v = v_0 + a t \). Here, \( v_0 = 26 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \), \( t = 2 \, \text{s} \).

Substitute: \( v = 26 - 10 \cdot 2 = 26 - 20 = 6 \, \text{m/s} \).

The velocity is \( 6 \, \text{m/s} \).

4 m/s
6 m/s
8 m/s
10 m/s
2

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