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22. (I) A car slows down from 28 m/s to rest in a distance of

88 m. What was its acceleration, assumed constant?

User Kvz
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1 Answer

4 votes

Answer:

a = -8.912 m/s²

Step-by-step explanation:

Given,

The initial velocity of the car, u = 28 m/s

The final velocity of the car, v = 0

The distance traveled by car, d = 88 m

The velocity displacement relation is given by the formula

v = d/t

∴ t = d/v

Substituting in the above values in the given equation

t = 88/28

= 3.142 s

The acceleration is given by the formula

a = (v-u)/t

= (0 - 28)/3.142

= -8.912 m/s²

The negative sign is that the car is decelerating.

Hence, acceleration a = -8.912 m/s²

User Krish KvR
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