9th Class Physics Unit 2 Kinematics Numericals with Solutions

2.2.A car is moving with an average speed of 72 km h^{-1}. How much time will it take to cover a distance of 360 km?

Given

Speed v = 72 km h^-1

Distance S = 360 km

Formula

Time t = Sv

Working

t = 36072

Result

Time t = 5 h

2.3.A truck starts from rest. It reaches a velocity of 90 km h^{-1} in 50 seconds. Find its average acceleration.

Given

Initial velocity v_i = 0 ms^-1

Final velocity v_f = 90 kmh^-1 = 90 × 10003600 = 25 ms^-1

Time t = 50 s

Formula

Average acceleration a_av = v_f - v_it

Working

a_av = 25 - 050

a_av = 2550

Result

Acceleration a_av = 0.5 ms^-2

2.4.A car passes a green traffic signal while moving with a velocity of 5m s^{-1}. It then accelerates to 1.5m s^{-2}. What is the velocity of car after 5 seconds?

Given

Initial velocity v_i = 5 ms^-1

Acceleration a_ac = 1.5 ms^-2

Time t = 5 s

Formula

Final velocity v_f = v_i + a_ac · t

Working

v_f = 5 + (1.5)(5)

v_f = 5 + 7.5

Result

Final velocity v_f = 12.5 ms^-1

2.5.A motorcycle initially travelling at 18 km h^{-1} accelerates at constant rate of 2m s^{-2}. How far will the motorcycle go in 10 seconds?

Given

Initial velocity v_i = 18 kmh^-1 = 18 × 10003600 = 5 ms^-1

Acceleration a = 2 ms^-2

Time t = 10 s

Formula

Distance S = v_i t + 12at²

Working

S = (5)(10) + 12(2)(10)²

S = 50 + 100

Result

Distance S = 150 m

2.6.A wagon is moving on the road with a velocity of 54km h^{-1}. Brakes are applied suddenly. The wagon covers a distance of 25 m before stopping. Determine the acceleration of the wagon.

Given

Initial velocity v_i = 54 kmh^-1 = 54 × 10003600 = 15 ms^-1

Final velocity v_f = 0 ms^-1

Distance S = 25 m

Formula

S = v_i + v_f2 × t ⇒ t = 2Sv_i + v_f

Working

t = 2(25)15 + 0 = 5015

t = 3.3333 s

Formula

Acceleration a = v_f - v_it

Working

a = 0 - 153.3333

Result

Acceleration a = -4.5 ms^-2

2.7.A stone is dropped from a height of 45 m. How long will it take to reach the ground? What will be its velocity just before hitting the ground?

Given

Height h = 45 m

Initial velocity v_i = 0 ms^-1

Gravitational acceleration g = 10 ms^-2

Formula

Using 2nd equation of motion h = v_i t + 12gt²

Working

45 = (0)t + 12(10)t²

45 = 5t²

t² = 9

Result

Time t = 3 s

Formula

Using 3rd equation of motion 2gh = v_f² - v_i²

Working

2(10)(45) = v_f² - (0)²

900 = v_f²

Result

Final velocity v_f = 30 ms^-1

2.8.A car travels 10 km with an average velocity of 20m s^{-1}. Then it travels in the same direction through a diversion at an average velocity of 4m s^{-1} for the next 0.8 km. Determine the average velocity of the car for the total journey.

Given

Distance 1 S₁ = 10 km = 10,000 m

Velocity 1 v₁ = 20 ms^-1

Distance 2 S₂ = 0.8 km = 800 m

Velocity 2 v₂ = 4 ms^-1

Formula

S = S₁ + S₂

S = 10,000 + 800 = 10,800 m

t₁ = S₁v₁

Working

t₁ = 10,00020 = 500 s

Formula

t₂ = S₂v₂

Working

t₂ = 8004 = 200 s

t = 500 + 200 = 700 s

Formula

Average velocity V_av = St

Working

V_av = 10,800700

Result

Average velocity V_av = 15.4 ms^-1

2.9.A ball is dropped from the top of a tower. The ball reaches the ground in 5 seconds. Find the height of the tower and the velocity of the ball with which it strikes the ground.

Given

Time t = 5 s

Acceleration due to gravity g = 10 ms^-2

Initial velocity v_i = 0 ms^-1

Formula

Using 2nd equation of motion h = v_i t + 12gt²

Working

h = (0)(5) + 12(10)(5)²

h = 0 + (5)(25)

Result

Height h = 125 m

Formula

Using 1st equation of motion v_f = v_i + gt

Working

v_f = (0) + (10)(5)

v_f = 0 + 50

Result

Final velocity v_f = 50 ms^-1

2.10.A cricket ball is hit so that it travels straight up in the air. An observer notes that it took 3 seconds to reach the highest point. What was the initial velocity of the ball? If the ball was hit 1 m above the ground, how high did it rise from the ground?

Given

Time to highest point t = 3 s

Gravitational acceleration g = -10 ms^-2

Final velocity at highest point v_f = 0 ms^-1

Height above ground h_above = 1 m

Formula

Using 1st equation of motion v_f = v_i + gt

Working

0 = v_i + (-10)(3)

0 = v_i - 30

Result

Initial velocity v_i = 30 ms^-1

Formula

Using 2nd equation of motion h = v_i t + 12gt²

Working

h = (30)(3) + 12(-10)(3)²

h = 90 - 5(9)

h = 90 - 45 = 45 m

Total height from ground Total height = 45 + 1 = 46 m

Unit 2: Kinematics — Numericals