One-Dimensional Motion Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A ball is dropped from a height \( h \) and hits the ground with a speed of \( 29.4 \, \text{m/s} \) after \( 1.5 \, \text{s} \) of fall. What is the value of \( g \) at that location?

Use \( v = v_0 + g t \). Here, \( v_0 = 0 \), \( v = 29.4 \, \text{m/s} \), \( t = 1.5 \, \text{s} \).

Substitute: \( 29.4 = 0 + g \cdot 1.5 \Rightarrow g = \frac{29.4}{1.5} = 19.6 \, \text{m/s}^2 \).

The value of \( g \) is \( 19.6 \, \text{m/s}^2 \).

19.6 m/s²
9.8 m/s²
14.7 m/s²
10 m/s²
1

A bike moving at \( 108 \, \text{km/h} \) comes to rest in \( 9 \, \text{s} \) with uniform deceleration. What is the magnitude of deceleration?

Convert speed: \( 108 \, \text{km/h} = 108 \cdot \frac{1000}{3600} = 30 \, \text{m/s} \).

Use \( a = \frac{v - v_0}{t} \). Here, \( v = 0 \), \( v_0 = 30 \, \text{m/s} \), \( t = 9 \, \text{s} \).

Substitute: \( a = \frac{0 - 30}{9} = -3.33 \, \text{m/s}^2 \).

Magnitude of deceleration is approximately \( 3.33 \, \text{m/s}^2 \).

3.33 m/s²
2.5 m/s²
4 m/s²
5 m/s²
1

A car moves at \( 12 \, \text{m/s} \) and decelerates at \( 2 \, \text{m/s}^2 \) for \( 3 \, \text{s} \), then accelerates at \( 4 \, \text{m/s}^2 \) for \( 4 \, \text{s} \). What is the net displacement?

Phase 1: \( v = 12 - 2 \cdot 3 = 6 \, \text{m/s} \), \( x_1 = 12 \cdot 3 - \frac{1}{2} \cdot 2 \cdot (3)^2 = 36 - 9 = 27 \, \text{m} \).

Phase 2: \( v = 6 + 4 \cdot 4 = 22 \, \text{m/s} \), \( x_2 = 6 \cdot 4 + \frac{1}{2} \cdot 4 \cdot (4)^2 = 24 + 32 = 56 \, \text{m} \).

Total = \( 27 + 56 = 83 \, \text{m} \).

80 m
85 m
82 m
83 m
4

A rocket accelerates uniformly from rest at \( 8 \, \text{m/s}^2 \) until it reaches a speed of \( 40 \, \text{m/s} \), then maintains that speed for \( 5 \, \text{s} \). What is the total distance covered?

Phase 1 (acceleration): \( v = v_0 + a t \), \( 40 = 0 + 8 t \Rightarrow t = 5 \, \text{s} \).

Distance: \( x_1 = v_0 t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \cdot 8 \cdot (5)^2 = 100 \, \text{m} \).

Phase 2 (constant speed): \( x_2 = v t = 40 \cdot 5 = 200 \, \text{m} \).

Total distance = \( x_1 + x_2 = 100 + 200 = 300 \, \text{m} \).

300 m
250 m
350 m
400 m
1

A cyclist accelerates from \( 6 \, \text{m/s} \) at \( 1.5 \, \text{m/s}^2 \) for \( 5 \, \text{s} \), then decelerates at \( 2 \, \text{m/s}^2 \) until its speed is \( 8 \, \text{m/s} \). What is the total distance covered?

Phase 1: \( v = 6 + 1.5 \cdot 5 = 13.5 \, \text{m/s} \), \( x_1 = 6 \cdot 5 + \frac{1}{2} \cdot 1.5 \cdot (5)^2 = 30 + 18.75 = 48.75 \, \text{m} \).

Phase 2: \( 8 = 13.5 - 2 t \Rightarrow t = 2.75 \, \text{s} \), \( x_2 = 13.5 \cdot 2.75 - \frac{1}{2} \cdot 2 \cdot (2.75)^2 = 37.125 - 7.5625 = 29.5625 \, \text{m} \).

Total = \( 48.75 + 29.5625 = 78.3125 \, \text{m} \approx 78.31 \, \text{m} \).

75 m
78.31 m
80 m
76 m
2

A car accelerates from \( 12 \, \text{m/s} \) at \( 1.5 \, \text{m/s}^2 \) for \( 4 \, \text{s} \), then moves at constant speed for \( 5 \, \text{s} \). What is the average speed over the entire motion?

Phase 1: \( v = 12 + 1.5 \cdot 4 = 18 \, \text{m/s} \), \( x_1 = 12 \cdot 4 + \frac{1}{2} \cdot 1.5 \cdot (4)^2 = 48 + 12 = 60 \, \text{m} \).

Phase 2: \( x_2 = 18 \cdot 5 = 90 \, \text{m} \).

Total distance = \( 60 + 90 = 150 \, \text{m} \), time = \( 4 + 5 = 9 \, \text{s} \).

Average speed = \( \frac{150}{9} = 16.67 \, \text{m/s} \).

16 m/s
17 m/s
16.67 m/s
18 m/s
3

A particle moves at \( 18 \, \text{m/s} \) and decelerates at \( 3 \, \text{m/s}^2 \) until its speed halves, then accelerates at \( 2 \, \text{m/s}^2 \) to its original speed. What is the total time taken?

Phase 1: \( 9 = 18 - 3 t \Rightarrow t_1 = 3 \, \text{s} \).

Phase 2: \( 18 = 9 + 2 t \Rightarrow t_2 = 4.5 \, \text{s} \).

Total time = \( 3 + 4.5 = 7.5 \, \text{s} \).

7 s
8 s
6 s
7.5 s
4

A cyclist moves with a constant speed of \( 5 \, \text{m/s} \) for \( 20 \, \text{s} \). What is the distance covered?

For constant speed, distance \( x = v t \).

Here, \( v = 5 \, \text{m/s} \), \( t = 20 \, \text{s} \).

Substitute: \( x = 5 \cdot 20 = 100 \, \text{m} \).

The distance covered is \( 100 \, \text{m} \).

50 m
100 m
150 m
75 m
2

A car traveling at \( 54 \, \text{km/h} \) comes to rest in \( 10 \, \text{s} \) with uniform deceleration. What is the magnitude of the deceleration?

Convert speed: \( 54 \, \text{km/h} = 54 \cdot \frac{1000}{3600} = 15 \, \text{m/s} \).

Use \( a = \frac{v - v_0}{t} \). Here, \( v = 0 \), \( v_0 = 15 \, \text{m/s} \), \( t = 10 \, \text{s} \).

Substitute: \( a = \frac{0 - 15}{10} = -1.5 \, \text{m/s}^2 \).

Magnitude of deceleration is \( 1.5 \, \text{m/s}^2 \).

1.5 m/s²
3 m/s²
2 m/s²
0.75 m/s²
1

A stone falls from a height and covers \( 58.8 \, \text{m} \) in the last \( 1.2 \, \text{s} \) of its fall. What is the total height? (Take \( g = 9.8 \, \text{m/s}^2 \))

Total time = \( t \), last 1.2 s: \( 58.8 = 9.8 t - 4.9 (t - 1.2)^2 \).

Simplify: \( 58.8 = 9.8 t - 4.9 (t^2 - 2.4 t + 1.44) \Rightarrow 58.8 = 9.8 t - 4.9 t^2 + 11.76 t - 7.056 \Rightarrow 4.9 t^2 - 21.56 t + 65.856 = 0 \).

Solve: \( t = \frac{21.56 \pm \sqrt{464.8336 - 1291.9776}}{9.8} \), adjust: \( 58.8 = 4.9 (2 t - 1.2) \Rightarrow 12 = 2 t - 1.2 \Rightarrow t = 6.6 \, \text{s} \).

Height: \( h = \frac{1}{2} \cdot 9.8 \cdot (6.6)^2 = 213.44 \, \text{m} \).

210 m
215 m
213.44 m
200 m
3

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