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# position and movement

In an emergency,a car moving at 50km/h cna be stopped in 15m. What is the stopped distance of the car if the speed is 70km/h,assume the same deceleration?

A.7.7m

B.10.9m

C.17.7m

D.21.0m

E.29.4m

唔該列埋算式

### 1 Answer

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- Prof. PhysicsLv 71 decade agoFavorite Answer
Initial velocity, u = 50 km/h = 50 / 3.6 = 13.89 m/s

Final velocity, v = 0

Stopping distance, s = 15 m

By equation of motion,

v2 = u2 - 2as

0 = (13.89)2 - 2a(15)

Deceleration, a = 6.43 ms-2

Now, initial velocity, u = 70 km/h = 70/3.6 = 19.44 m/s

By equation of motion,

v2 = u2 - 2as

0 = (19.44)2 - 2(6.43)s

Stopping distance, s = 29.4 m

The answer is E.

Source(s): Physics king

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