Kinematics

Practice Questions:

Question 101: moderate

A taxi leaves the station X for station Y every 10 minutes. Simultaneously, a taxi leaves the station Y also for station X every 10 minutes. The taxis move at the same constant speed and go from X to Y or vice-versa in 2 hours. How many taxis coming from the other side will each taxi meet enroute from Y to X ?

1. 10

2. 11

3. 12

4. 23

View Answer

Each taxi takes 2 hours, and taxis leave every 10 minutes.

Taxis already on the route from X to Y:

$\frac{120}{10}$ $= 12$
Taxis that start after from X to Y:

$\frac{120}{10}$ $= 12$
Exclude the simultaneously departing taxi:

$12 + 11 = 23$

Final Answer: 23 taxis.

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Question 102: easy

A car travelling with a velocity of 90 km/h slowed down to 54 km/h in 15 s. The retardation is

1. 0.67 m/s²

2. 1 m/s²

3. 1.25 m/s²

4. 1.5 m/s²

View Answer

Using the formula

$a = \frac{v_f - v_i}{t}$

Converting to m/s:

$90$

km/h =

$25$

m/s,

$54$

km/h =

$15$

m/s.

$a = \frac{15 - 25}{15} = \frac{-10}{15} = -0.67 \text{ m/s}^2$

Retardation = 0.67 m/s².

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Question 103: easy

A person driving a car with a speed 54 km/h, suddenly sees a boy crossing the road. If the distance moved by car, before the person applies brakes is 5 m, the reaction time of the person is

1. 0.5 sec

2. 0.66 sec

3. 0.33 sec

4. 1 sec

View Answer

Using

$t = \frac{s}{v}$

, with

$v = 54$

km/h = 15 m/s and

$s = 5$

$t = \frac{5}{15} = 0.33 \text{ s}$

Reaction time = 0.33 s.

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Question 104: easy

When a car is stopped by applying brakes, it stops after travelling a distance of 80 m. If speed of car is halved and same retarding acceleration is applied then it stops after travelling a distance of

1. 20 m

2. 50 m

3. 75 m

4. 100 m

View Answer

Using the stopping distance formula:

$s \propto u^2$

If speed is halved, new stopping distance:

$s_2 = \frac{100}{4} = 25 \text{ m}$

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Question 105: moderate

Two cars A and B are approaching each other head-on with speeds 20 m/s and 10 m/s respectively. When their separation is X then A and
B start braking at 4 m/s² and 2m/s² respectively. Minimum value of X to avoid collision is

1. 60 m

2. 75 m

3. 80 m

4. 90 m

View Answer

Using the stopping distance formula:

$s = \frac{u^2}{2a}$

For Car A (

$u_A = 20$

m/s,

$a_A = 4$

m/s²):

$s_A = \frac{20^2}{2 \times 4} = \frac{400}{8} = 50 \text{ m}$

For Car B (

$u_B = 10$

m/s,

$a_B = 2$

m/s²):

$s_B = \frac{10^2}{2 \times 2} = \frac{100}{4} = 25 \text{ m}$

Total minimum separation to avoid collision:

$X = s_A + s_B = 50 + 25 = 75 \text{ m}$

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Question 106: difficult

Two bullets are fired horizontally and simultaneously towards each other from roof tops of two buildings 100 m apart and of same height of 200 m, with the same velocity of 25 m/s. When and where will the two bullets collide? (g = 10 m/s²)

1. They will not collide

2. After 2 s at a height of 180 m

3. After 2 s at a height of 20 m

4. After 4 s at a height of 120 m

View Answer

Horizontal motion: Bullets move towards each other with relative velocity $50$

m/s. Time to meet:

$t = \frac{100}{50} = 2 \text{ s}$
Vertical motion: Free fall equation:
$H = 200 - 5t^2 = 200 - 5(2)^2 = 180 \text{ m}$

Answer: (2) After 2 s at a height of 180 m.

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Question 107: easy

A ball is thrown horizontally from the top of a 80 m tall building with a speed of 20 m/s. Find the time it takes to reach the ground

1. 4 sec

2. 5 sec

3. 6 sec

4. 7 sec

View Answer

Using formula for Time:

$t = \sqrt{\frac{2h}{g}}$

$t = g 2 h$

$t = \sqrt{\frac{2 \times 80}{10}} = \sqrt{16} = 4 \text{ s}$

$t = 102 \times 80 = 16 = 4 s$

Correct Answer: (1) 4 s

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Question 108: easy

A ball is thrown horizontally with a speed of 20 m/s from a height of 80 m. Find the horizontal distance before hitting the ground.

1. 60 m

2. 70 m

3. 80 m

4. 90 m

View Answer

Using the formula for range in horizontal projectile motion:

$R = v_x \times \sqrt{\frac{2h}{g}}$

$R = v_{x} \times g 2 h$

$R = 20 \times \sqrt{\frac{2 \times 80}{10}}$

$R = 20 \times 102 \times 80$

$R = 20 \times \sqrt{16} = 20 \times 4 = 80 \text{ m}$

$R = 20 \times 16 = 20 \times 4 = 80 m$

Correct Answer: (3) 80 m

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Question 109: easy

A helicopter flying at a height of 125 m and moving horizontally with a speed of 10 m/s drops a package. Find the horizontal distance it covers before hitting the ground.

1. 20 m

2. 30 m

3. 40 m

4. 50 m

View Answer

Using the direct range formula for a projectile dropped from a height:

$R = v_x \times \sqrt{\frac{2h}{g}}$

$R = 10 \times \sqrt{\frac{2 \times 125}{10}}$

$R = 10 \times \sqrt{25} = 10 \times 5 = 50 \text{ m}$

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Question 110: easy

A helicopter is moving horizontally at a constant speed when it drops a package from a certain height. What will be the observed path of the package when viewed from the ground ?

1. A straight vertical line

2. A straight horizontal line

3. A parabola

4. A circle

View Answer

\[ y = -\frac{g}{2 v_x^2} x^2 \]

This is of the form

$y = ax^2$

, which represents a parabolic path. Hence, the object follows a parabolic trajectory when viewed from the ground.

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