Question

A car of mass 1000kg travel a distance of 120m/s before coming to rest what is the force exerted by the breaks to stop it
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Answer 1

Answer:

Acceleration due to friction (retardation) (a) = - μg=−0.9×10=−9m/s2

Final velocity, v=0m/s

Initial velocity, u=20m/s

Applying formula: v2=u2+2as

Where s is the distance traveled by car before it come to rest.

⇒02=202+2(−9)s

⇒18s=400

⇒s=22.2 m