Kinematics One Dimensional Motion Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A car accelerates from \( 5 \, \text{m/s} \) at \( 2 \, \text{m/s}^2 \) for \( 6 \, \text{s} \), then moves at constant speed for \( 3 \, \text{s} \). What is the average speed over this time?

Phase 1: \( v = 5 + 2 \cdot 6 = 17 \, \text{m/s} \), \( x_1 = 5 \cdot 6 + \frac{1}{2} \cdot 2 \cdot (6)^2 = 30 + 36 = 66 \, \text{m} \).

Phase 2: \( x_2 = 17 \cdot 3 = 51 \, \text{m} \).

Total distance = \( 66 + 51 = 117 \, \text{m} \), time = \( 6 + 3 = 9 \, \text{s} \).

Average speed = \( \frac{117}{9} = 13 \, \text{m/s} \).

12 m/s
14 m/s
13 m/s
15 m/s
3

A car moving at \( 72 \, \text{km/h} \) is brought to rest in \( 4 \, \text{s} \) by applying brakes. What is the retardation?

Convert speed: \( 72 \, \text{km/h} = 72 \cdot \frac{1000}{3600} = 20 \, \text{m/s} \).

Use \( a = \frac{v - v_0}{t} \). Here, \( v = 0 \), \( v_0 = 20 \, \text{m/s} \), \( t = 4 \, \text{s} \).

Substitute: \( a = \frac{0 - 20}{4} = -5 \, \text{m/s}^2 \).

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

2.5 m/s²
5 m/s²
10 m/s²
7.5 m/s²
2

A ball is dropped from rest. What is its velocity after falling for \( 1.5 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v = v_0 + g t \). Here, \( v_0 = 0 \), \( g = 10 \, \text{m/s}^2 \), \( t = 1.5 \, \text{s} \).

Substitute: \( v = 0 + 10 \cdot 1.5 = 15 \, \text{m/s} \).

The velocity is \( 15 \, \text{m/s} \).

10 m/s
15 m/s
20 m/s
5 m/s
2

A car moving at \( 22 \, \text{m/s} \) decelerates uniformly to rest in \( 5 \, \text{s} \). What is the distance covered?

Find \( a = \frac{v - v_0}{t} = \frac{0 - 22}{5} = -4.4 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 22 \cdot 5 + \frac{1}{2} (-4.4) (5)^2 = 110 - 55 = 55 \, \text{m} \).

The distance covered is \( 55 \, \text{m} \).

50 m
60 m
55 m
44 m
3

A train starts from rest and accelerates uniformly at \( 0.5 \, \text{m/s}^2 \) for \( 20 \, \text{s} \). What is the final velocity?

Use the equation \( v = v_0 + a t \).

Here, initial velocity \( v_0 = 0 \), acceleration \( a = 0.5 \, \text{m/s}^2 \), time \( t = 20 \, \text{s} \).

Substitute: \( v = 0 + 0.5 \cdot 20 = 10 \, \text{m/s} \).

The final velocity is \( 10 \, \text{m/s} \).

10 m/s
15 m/s
5 m/s
20 m/s
1

A ball is thrown upwards with \( 25 \, \text{m/s} \) from the ground. 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, \( v = 0 \), \( v_0 = 25 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = (25)^2 + 2 (-10) h \Rightarrow 0 = 625 - 20 h \Rightarrow h = \frac{625}{20} = 31.25 \, \text{m} \).

The maximum height is \( 31.25 \, \text{m} \).

25 m
50 m
31.25 m
62.5 m
3

A cyclist travels at a constant speed of \( 18 \, \text{km/h} \) for \( 10 \, \text{minutes} \). What is the distance covered?

Convert speed: \( 18 \, \text{km/h} = 18 \cdot \frac{1000}{3600} = 5 \, \text{m/s} \).

Time = \( 10 \, \text{min} = 10 \cdot 60 = 600 \, \text{s} \).

Distance \( x = v t = 5 \cdot 600 = 3000 \, \text{m} = 3 \, \text{km} \).

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

1.5 km
2 km
2.5 km
3 km
4

A parachutist falls freely for \( 2.5 \, \text{s} \) before opening the parachute, then decelerates at \( 4 \, \text{m/s}^2 \) to hit the ground at \( 5 \, \text{m/s} \). What is the total time of descent? (Take \( g = 10 \, \text{m/s}^2 \))

Free fall: \( v = g t = 10 \cdot 2.5 = 25 \, \text{m/s} \).

Deceleration phase: \( v = v_0 + a t \), \( 5 = 25 - 4 t \Rightarrow 4 t = 20 \Rightarrow t = 5 \, \text{s} \).

Total time = \( 2.5 + 5 = 7.5 \, \text{s} \).

6 s
7.5 s
8 s
7 s
2

A train moving at \( 90 \, \text{km/h} \) accelerates at \( 0.8 \, \text{m/s}^2 \) for \( 15 \, \text{s} \), then decelerates at \( 2.5 \, \text{m/s}^2 \) to rest. What is the total distance covered?

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

Phase 1: \( v = 25 + 0.8 \cdot 15 = 37 \, \text{m/s} \), \( x_1 = 25 \cdot 15 + \frac{1}{2} \cdot 0.8 \cdot (15)^2 = 375 + 90 = 465 \, \text{m} \).

Phase 2: \( t = \frac{37}{2.5} = 14.8 \, \text{s} \), \( x_2 = 37 \cdot 14.8 - \frac{1}{2} \cdot 2.5 \cdot (14.8)^2 = 547.6 - 273.8 = 273.8 \, \text{m} \).

Total = \( 465 + 273.8 = 738.8 \, \text{m} \).

730 m
750 m
738.8 m
720 m
3

A ball is thrown vertically upwards with an initial speed of \( 28 \, \text{m/s} \). What is the total time of flight? (Take \( g = 10 \, \text{m/s}^2 \))

Time to max height: \( v = v_0 + a t \), \( 0 = 28 - 10 t \Rightarrow t = 2.8 \, \text{s} \).

Total time = time up + time down = \( 2.8 \, \text{s} + 2.8 \, \text{s} = 5.6 \, \text{s} \).

The total time is \( 5.6 \, \text{s} \).

5.6 s
2.8 s
4 s
6 s
1

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