One-Dimensional Motion Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A train moving at \( 72 \, \text{km/h} \) passes a pole in \( 9 \, \text{s} \). If it decelerates uniformly to rest in \( 30 \, \text{s} \) after passing the pole, what is the length of the train?

Speed: \( 72 \, \text{km/h} = 20 \, \text{m/s} \).

Length = distance to pass pole = \( 20 \cdot 9 = 180 \, \text{m} \).

Check: Deceleration \( a = \frac{0 - 20}{30} = -\frac{2}{3} \, \text{m/s}^2 \). Distance to stop = \( 20 \cdot 30 + \frac{1}{2} (-\frac{2}{3}) (30)^2 = 600 - 300 = 300 \, \text{m} \).

Length = \( 180 \, \text{m} \) (initial pass distance).

180 m
200 m
150 m
240 m
1

A stone is dropped from a height of \( 45 \, \text{m} \) on a planet where it takes \( 3 \, \text{s} \) to reach the ground. What is the acceleration due to gravity on this planet?

\( h = \frac{1}{2} g t^2 \), \( 45 = \frac{1}{2} g (3)^2 \Rightarrow 45 = 4.5 g \Rightarrow g = 10 \, \text{m/s}^2 \).

9 m/s²
11 m/s²
8 m/s²
10 m/s²
4

A bike accelerates uniformly from rest to \( 20 \, \text{m/s} \) over a distance of \( 40 \, \text{m} \). What is its acceleration?

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

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

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

2.5 m/s²
10 m/s²
5 m/s²
7.5 m/s²
3

A stone falls freely from rest and reaches a velocity of \( 14.7 \, \text{m/s} \). How long did it fall? (Take \( g = 9.8 \, \text{m/s}^2 \))

Use \( v = v_0 + g t \). Here, \( v_0 = 0 \), \( v = 14.7 \, \text{m/s} \), \( g = 9.8 \, \text{m/s}^2 \).

Substitute: \( 14.7 = 0 + 9.8 t \Rightarrow t = \frac{14.7}{9.8} = 1.5 \, \text{s} \).

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

1 s
2 s
1.5 s
3 s
3

Two stones are dropped from a height of \( 125 \, \text{m} \), with a \( 2 \, \text{s} \) interval between them. How far apart are they when the second stone has fallen for \( 3 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Stone 1 (after 5 s): \( y_1 = \frac{1}{2} \cdot 10 \cdot (5)^2 = 125 \, \text{m} \) (hits ground).

Stone 2 (after 3 s): \( y_2 = \frac{1}{2} \cdot 10 \cdot (3)^2 = 45 \, \text{m} \).

Distance apart = \( 125 - 45 = 80 \, \text{m} \).

70 m
90 m
80 m
100 m
3

A ball is thrown upwards with a speed of \( 38 \, \text{m/s} \). What is its velocity after \( 3 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v = v_0 + a t \). Here, \( v_0 = 38 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \), \( t = 3 \, \text{s} \).

Substitute: \( v = 38 - 10 \cdot 3 = 38 - 30 = 8 \, \text{m/s} \).

The velocity is \( 8 \, \text{m/s} \).

5 m/s
8 m/s
10 m/s
12 m/s
2

A rocket ascends with an acceleration of \( 4 \, \text{m/s}^2 \) for \( 6 \, \text{s} \), then its engines shut off. How long does it take to reach its maximum height from the start? (Take \( g = 10 \, \text{m/s}^2 \))

After 6 s: \( v = 4 \cdot 6 = 24 \, \text{m/s} \).

Deceleration phase: \( v = 24 - 10 t \), \( 0 = 24 - 10 t \Rightarrow t = 2.4 \, \text{s} \).

Total time = \( 6 + 2.4 = 8.4 \, \text{s} \).

8 s
8.4 s
9 s
7.5 s
2

Two stones are dropped from a height of \( 80 \, \text{m} \), \( 1.5 \, \text{s} \) apart. How long after the second stone is released do they meet if the first rebounds at \( 15 \, \text{m/s} \) upon hitting the ground? (Take \( g = 10 \, \text{m/s}^2 \))

First stone: \( 80 = \frac{1}{2} \cdot 10 \cdot t^2 \Rightarrow t = 4 \, \text{s} \).

Second stone released at \( t = 2.5 \, \text{s} \), meets after \( t' \) s.

First stone after rebound: \( y_1 = 15 (t' + 1.5) \), second stone: \( y_2 = 5 t'^2 \).

\( 80 - y_1 = y_2 \Rightarrow 80 - 15 (t' + 1.5) = 5 t'^2 \Rightarrow 80 - 15 t' - 22.5 = 5 t'^2 \Rightarrow 5 t'^2 + 15 t' - 57.5 = 0 \).

Solve: \( t = \frac{-15 \pm \sqrt{225 + 1150}}{10} = \frac{-15 \pm 37.08}{10} \), \( t = 2.21 \, \text{s} \) (positive root).

2.21 s
2.5 s
2 s
3 s
1

A stone is thrown upwards at \( 30 \, \text{m/s} \) from a \( 55 \, \text{m} \) tower. How far below the tower’s top is it after \( 7 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Displacement: \( y = 30 \cdot 7 - \frac{1}{2} \cdot 10 \cdot (7)^2 = 210 - 245 = -35 \, \text{m} \).

Distance below top = \( 35 \, \text{m} \).

35 m
40 m
30 m
45 m
1

A ball is dropped from rest. What is the distance covered in the 3rd second of its fall? (Take \( g = 10 \, \text{m/s}^2 \))

Distance in \( n \)-th second = \( g \cdot \frac{2n - 1}{2} \). For 3rd second, \( n = 3 \).

Substitute: \( d = 10 \cdot \frac{2 \cdot 3 - 1}{2} = 10 \cdot \frac{5}{2} = 25 \, \text{m} \).

The distance covered is \( 25 \, \text{m} \).

15 m
20 m
30 m
25 m
4

Performance Summary

Score:

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