Question

A ball starting from the rest hits the wall with the velocity 40 m/s. Find the distance covered by the ball in 4 seconds​

Answer 1

Answer:

velocity after 2 sec.

v=40+(−10)t

=40−10∗2=20m/s

Explanation:

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Answer 2

Given :-

• A ball starting from the rest hits the wall with the velocity 40 m/s.

To find :-

• Distance covered by the ball in 4 seconds

Solution :-

• Initial velocity (u) = 0 (started from rest)

• Final velocity (v) = 40m/s

• Time (t) = 4 s

According to first equation of motion

→ v = u + at

→ 40 = 0 + a × 4

→ 40 = 4a

→ a = 40/4

→ a = 10 m/s²

Now, according to second equation of motion

→ s = ut + ½ at²

→ s = 0 × 4 + ½ × 10 × (4)²

→ s = ½ × 10 × 16

→ s = 5 × 16

→ s = 80 m

Hence,

• Distance covered by ball is 80 m

More to know :-

• S.I unit of speed is m/s
• v² - u² = 2as (Third equation of motion)
• The rate of change in velocity is known as acceleration
• Retardation of body → When the value of acceleration is negative.