Question

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?

Answer 1

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.