One-Dimensional Motion Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A rocket accelerates vertically at \( 6 \, \text{m/s}^2 \) from rest for \( 4 \, \text{s} \), then its engines fail. What is the maximum height reached above the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Phase 1: \( v = 6 \cdot 4 = 24 \, \text{m/s} \), \( h_1 = \frac{1}{2} \cdot 6 \cdot (4)^2 = 48 \, \text{m} \).

Phase 2: \( v^2 = v_0^2 - 2 g h \), \( 0 = (24)^2 - 2 \cdot 10 \cdot h_2 \Rightarrow h_2 = \frac{576}{20} = 28.8 \, \text{m} \).

Total height = \( 48 + 28.8 = 76.8 \, \text{m} \).

75 m
76.8 m
80 m
70 m
2

A helicopter ascends vertically at \( 6 \, \text{m/s} \) and releases a package after \( 10 \, \text{s} \). How long after release does the package hit the ground if it was initially \( 50 \, \text{m} \) above the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Height at release: \( h = 50 + 6 \cdot 10 = 110 \, \text{m} \).

Initial velocity of package = \( 6 \, \text{m/s} \) upward.

Use \( y = v_0 t - \frac{1}{2} g t^2 \), ground level \( y = -110 \).

\( -110 = 6 t - 5 t^2 \Rightarrow 5 t^2 - 6 t - 110 = 0 \).

Solve: \( t = \frac{6 \pm \sqrt{36 + 2200}}{10} = \frac{6 \pm 47.6}{10} \), \( t = 5.36 \, \text{s} \) (positive root).

Time = \( 5.36 \, \text{s} \).

5.36 s
5 s
6 s
4.5 s
1

A car moves at \( 16 \, \text{m/s} \) and decelerates at \( 2.5 \, \text{m/s}^2 \) for \( 3 \, \text{s} \), then accelerates at \( 3.5 \, \text{m/s}^2 \) for \( 4 \, \text{s} \). What is the net displacement?

Phase 1: \( v = 16 - 2.5 \cdot 3 = 8.5 \, \text{m/s} \), \( x_1 = 16 \cdot 3 - \frac{1}{2} \cdot 2.5 \cdot (3)^2 = 48 - 11.25 = 36.75 \, \text{m} \).

Phase 2: \( v = 8.5 + 3.5 \cdot 4 = 22.5 \, \text{m/s} \), \( x_2 = 8.5 \cdot 4 + \frac{1}{2} \cdot 3.5 \cdot (4)^2 = 34 + 28 = 62 \, \text{m} \).

Total = \( 36.75 + 62 = 98.75 \, \text{m} \).

95 m
100 m
97 m
98.75 m
4

A rocket accelerates from rest at \( 6 \, \text{m/s}^2 \) for \( 5 \, \text{s} \), then moves at constant speed for \( 3 \, \text{s} \). What is the total distance covered?

Phase 1: \( v = 6 \cdot 5 = 30 \, \text{m/s} \), \( x_1 = \frac{1}{2} \cdot 6 \cdot (5)^2 = 75 \, \text{m} \).

Phase 2: \( x_2 = 30 \cdot 3 = 90 \, \text{m} \).

Total = \( 75 + 90 = 165 \, \text{m} \).

150 m
170 m
160 m
165 m
4

A bus moves at a constant speed of \( 36 \, \text{km/h} \) for \( 30 \, \text{minutes} \). What is the distance covered?

Convert speed: \( 36 \, \text{km/h} = 36 \cdot \frac{1000}{3600} = 10 \, \text{m/s} \).

Time = \( 30 \, \text{min} = 30 \cdot 60 = 1800 \, \text{s} \).

Distance \( x = v t = 10 \cdot 1800 = 18000 \, \text{m} = 18 \, \text{km} \).

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

18 km
12 km
15 km
20 km
1

A particle starts from rest and moves with a uniform acceleration of \( 3 \, \text{m/s}^2 \) for \( 6 \, \text{s} \). What is the distance covered?

Use \( x = v_0 t + \frac{1}{2} a t^2 \).

Here, \( v_0 = 0 \), \( a = 3 \, \text{m/s}^2 \), \( t = 6 \, \text{s} \).

Substitute: \( x = 0 \cdot 6 + \frac{1}{2} \cdot 3 \cdot (6)^2 = 0 + 1.5 \cdot 36 = 54 \, \text{m} \).

The distance covered is \( 54 \, \text{m} \).

36 m
54 m
72 m
18 m
2

A ball is thrown upwards at \( 30 \, \text{m/s} \) from a \( 40 \, \text{m} \) tower. What is the time taken to reach a point \( 10 \, \text{m} \) above the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Displacement \( y = -30 \, \text{m} \), \( -30 = 30 t - 5 t^2 \Rightarrow 5 t^2 - 30 t - 30 = 0 \Rightarrow t^2 - 6 t - 6 = 0 \).

Solve: \( t = \frac{6 \pm \sqrt{36 + 24}}{2} = \frac{6 \pm 7.75}{2} \), \( t = 6.875 \, \text{s} \) (downward path).

6.5 s
7 s
6.875 s
6 s
3

A ball is thrown vertically upwards with an initial speed of \( 15 \, \text{m/s} \). What is the time taken to reach its maximum height? (Take \( g = 10 \, \text{m/s}^2 \))

At maximum height, final velocity \( v = 0 \). Use \( v = v_0 + a t \).

Here, \( v_0 = 15 \, \text{m/s} \), \( a = -g = -10 \, \text{m/s}^2 \), \( v = 0 \).

Substitute: \( 0 = 15 - 10 t \Rightarrow 10 t = 15 \Rightarrow t = 1.5 \, \text{s} \).

The time taken is \( 1.5 \, \text{s} \).

1.5 s
3 s
2 s
1 s
1

A particle accelerates uniformly from rest to \( 12 \, \text{m/s} \) in \( 4 \, \text{s} \). What is the distance covered?

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

Use \( x = v_0 t + \frac{1}{2} a t^2 = 0 \cdot 4 + \frac{1}{2} \cdot 3 \cdot (4)^2 = 0 + 1.5 \cdot 16 = 24 \, \text{m} \).

The distance covered is \( 24 \, \text{m} \).

16 m
24 m
32 m
12 m
2

A body accelerates uniformly from rest to \( 24 \, \text{m/s} \) over a distance of \( 72 \, \text{m} \). What is the acceleration?

Use \( v^2 = v_0^2 + 2 a x \). Here, \( v_0 = 0 \), \( v = 24 \, \text{m/s} \), \( x = 72 \, \text{m} \).

Substitute: \( (24)^2 = 0 + 2 a (72) \Rightarrow 576 = 144 a \Rightarrow a = 4 \, \text{m/s}^2 \).

The acceleration is \( 4 \, \text{m/s}^2 \).

2 m/s²
6 m/s²
3 m/s²
4 m/s²
3

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0