Question

A car accelerates uniformly from 5m/s to 15m/s taking 7.5 seconds. How far did the car travel during this period?

Answer 1

A car accelerates uniformly from 5m/s to 10m/s in 5s. What distance was covered by the car?

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Initial velocity of the car (u)= 5 m/s

Final velocity of the car (v) = 10 m/s

Time taken for change in velocity (t)= 5 s

Therefore acceleration (a)= [10 m/s - 5 m/s]/ 5 s = 5 m/s /5s = 1 m/s²

Distance covered during these 5 seconds, is found using the relation

s = u t + ½ a t² = 5 m/s× 5 s + ½ 1 m/s² × 5² s²=25 m+ 12.5 m = 37.5 m

So car covers 37.5 m during the period its velocity changes from 5 m/s to 10 m/s.