Class 9 - Unit # 2 : Kinematics - Solved Numericals

Section 1

Types of Motion & Acceleration

Problem (d)

Calculate the acceleration of a bus that speeds up from 20 m/s to 40 m/s in 8 seconds. (Ans: 2.5 m/s²)

Given Data:

• Initial speed (v_i) = 20 m/s
• Final speed (v_f) = 40 m/s
• Time interval (t) = 8 s
• Acceleration (a) = ?

Solution:

According to the definition of acceleration:

a =

v_f - v_i

t

Substituting the given values into the formula:

a =

40 - 20

8

a =

20

8

a = 2.5 m/s²

Result:

The acceleration of the bus is 2.5 m/s².

Section 2

Equations of Motion

Problem (a)

A bus is moving on a road with 15 m/s and it accelerates at 5 m/s². Find the final velocity of the bus after 6 seconds. (Ans: 45 m/s)

Given Data:

• Initial velocity (v_i) = 15 m/s
• Acceleration (a) = 5 m/s²
• Time interval (t) = 6 s
• Final velocity (v_f) = ?

Solution:

Using the first equation of motion:

v_f = v_i + at

Substituting the values:

v_f = 15 + (5 × 6)

v_f = 15 + 30

v_f = 45 m/s

Result:

The final velocity of the bus is 45 m/s.

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Problem (b)

A car starts moving from rest with an acceleration of 5 m/s². Find out the time to travel 50 m distance. (Ans: 4.47 s)

Given Data:

• Initial velocity (v_i) = 0 m/s (starts from rest)
• Acceleration (a) = 5 m/s²
• Distance (S) = 50 m
• Time taken (t) = ?

Solution:

According to the second equation of motion:

S = v_it +

1

2

at²

Substituting the given values:

50 = (0 × t) +

1

2

× 5 × t²

50 = 2.5 × t²

Rearranging the equation to isolate t²:

t² =

50

2.5

= 20

Taking square root on both sides:

t = √20

t ≈ 4.47 s

Result:

The time to travel a 50 m distance is 4.47 seconds.

Section 3

Motion Due to Gravity

Problem (c)

A ball is dropped from a height of 50 m. What will be its velocity before touching the ground? (Ans: 31.3 m/s)

Given Data:

• Initial velocity (v_i) = 0 m/s (dropped from rest)
• Height (h) = 50 m
• Acceleration due to gravity (g) = 9.8 m/s²
• Final velocity (v_f) = ?

Solution:

Using the third equation of motion under gravity:

2gh = v_f² - v_i²

Substituting values:

2 × 9.8 × 50 = v_f² - (0)²

980 = v_f²

Taking square root on both sides:

v_f = √980

v_f ≈ 31.3 m/s

Result:

The final velocity before touching the ground is 31.3 m/s.

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Problem (d)

If a body is thrown upward with a vertical velocity of 50 m/s, calculate the maximum height which the body can reach. (Ans: 127.55 m)

Given Data:

• Initial velocity (v_i) = 50 m/s
• Final velocity (v_f) = 0 m/s (stops momentarily at maximum height)
• Acceleration due to gravity (g) = -9.8 m/s² (motion is upward against gravity)
• Maximum height (h) = ?

Solution:

Using the third equation of motion under gravity:

2gh = v_f² - v_i²

Substituting values:

2 × (-9.8) × h = (0)² - (50)²

-19.6 × h = -2500

Rearranging to solve for height (h):

h =

-2500

-19.6

h ≈ 127.55 m

Result:

The maximum height the body can reach is 127.55 m.

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Problem (e)

A ball falls down from a height of 70 m. How much time will the ball take to reach the ground? (Ans: 3.78 s)

Given Data:

• Initial velocity (v_i) = 0 m/s (released from rest)
• Height (h) = 70 m
• Acceleration due to gravity (g) = 9.8 m/s²
• Time taken (t) = ?

Solution:

Using the second equation of motion under gravity:

h = v_it +

1

2

gt²

Substituting the given values:

70 = (0 × t) +

1

2

× 9.8 × t²

70 = 4.9 × t²

Rearranging the equation to isolate t²:

t² =

70

4.9

≈ 14.286

Taking square root on both sides:

t = √14.286

t ≈ 3.78 s

Result:

The time to reach the ground is 3.78 seconds.
(Note: If calculated using g = 10 m/s², the time is exactly √14 ≈ 3.74 s)