? asked in 科學及數學其他 - 科學 · 1 decade ago

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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  • 1 decade ago
    Favorite 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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