A uniformly accelarating object covers a distance of 11 meters in 6.5 seconds. If it reaches a final velocity of 3.3 m/sec, what is its initial velocity?

1 Answer
Shwetank Mauria
Jan 21, 2017

Initial velocity is 0.0846 m/sec.

Explanation:

Some of the formulas used in uniformly accelerating motion with an acceleration a, initial velocity u - not known here, final velocity v - given here as 3.3 m/sec, time taken t is 6.5 sec and distance covered (in t sec) S, which is 11.0 meters, are

(A) v=u+at, (B) S=ut+1/2at^2 and (C) v^2-u^2=2aS

From (A) we can also have v-u=at and dividing (C) by this,

we get v+u=(2S)/t

As S=11.0 meters, v=3.3 m/sec and t=6.5 sec.

3.3+u=(2xx11)/6.5=22/6.5=44/13=3.3846

Hence u=3.3846-3.3=0.0846

i.e. initial velocity is 0.0846 m/sec.

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