Question

A scooterist on straight highway at 64 km/ h, sights an obstacle at distance of 50 m, and applies - 2.5 m/s^2 retardation will be able to avoid the accident.
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Answer 1

Answer :-

The scooterist will not be able to avoid the accident.

Explanation :-

We have :-

→ Initial velocity (u) = 64 km/h

→ Final velocity (v) = 0 m/s

→ Distance of obstacle = 50 m

→ Acceleration (a) = -2.5 m/s²

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Firstly, let's convert the initial velocity of the scooter from km/h to m/s.

⇒ 1 km/h = 5/18 m/s

⇒ 64 km/h = 64 × 5/18

⇒ 320/18 m/s

⇒ 17.78 m/s

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In order the avoid the accident, the stopping distance of the scooter should be less than 50 metre. Let's check that using 3rd equation of motion.

v² - u² = 2as

⇒ 0 - (17.78)² = 2(-2.5)s

⇒ -316.13 = -5s

⇒ s = -316.13/-5

⇒ s = 63.226 m

As 63.226 > 50, so the scooter will not be able to avoid the accident.

Answer 2

Given :-

A scooterist on straight highway at 64 km/ h, sights an obstacle at distance of 50 m, and applies - 2.5 m/s²

To Find :-

Whether he may avoid accident

Solution :-

1 km/h = 5/18 m/s

64 km/h = 64 × 5/18 = 320/18 m/s

Now

v² = u² + 2as

(0) = (320/18)² = 2(-2.5)(s)

0 - 102400/324 = -5s

-102400/324 = -5s

-102400/324 ÷ -5 = s

-102400/324 × 1/5 = s

20480/324 = s

63.2 = s

Now

Since 63.2 is greater than the given distance which is 50 m. So, he must have an accident.

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