Question

Motorbike running at 25 metre per second is slowed down to 5 metre per second in 2.5 seconds.

Calculate:

1)acceleration

2) distance covered in the time it slows down. ​

Answer 1

Answer :-

(1) Acceleration of the motorbike is -8 m/s² .

(2) Distance covered is 37.5 m .

Explanation :-

We have :-

→ Initial velocity (u) = 25 m/s

→ Final velocity (v) = 5 m/s

→ Time (t) = 2.5 seconds

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By using the 1st equation of motion, we get :-

v = u + at

⇒ 5 = 25 + a(2.5)

⇒ 5 - 25 = 2.5a

⇒ -20 = 2.5a

⇒ a = -20/2.5

⇒ a = -8 m/s²

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Now, by using the 2nd equation of motion :-

s = ut + ½at²

⇒ s = 25 × 2.5 + ½ × (-8) × (2.5)²

⇒ s = 62.5 + (-4)(6.25)

⇒ s = 62.5 - 25

⇒ s = 37.5 m

Answer 2

Answer:

Given :-

Motorbike running at 25 metre per second is slowed down to 5 metre per second in 2.5 seconds.

To Find :-

Acceleration

Distance covered

Solution :-

We know that

a = v - u/t

a = 5 - 25/2.5

a = -20/2.5

a = -8

Hence,

Acceleration = -8 m/s²

Now

s = ut + ½at²

s = (25)(2.5) + ½ × (-8) × (2.5)²

s = 62.5 + ½ × (-8) × 6.25

s = 62.5 + (-4) × 6.25

s = 62.5 - 25

s = 37.5 m