# 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?

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) $v = u + a t$, (B) $S = u t + \frac{1}{2} a {t}^{2}$ and (C) ${v}^{2} - {u}^{2} = 2 a S$

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

we get $v + u = \frac{2 S}{t}$

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

$3.3 + u = \frac{2 \times 11}{6.5} = \frac{22}{6.5} = \frac{44}{13} = 3.3846$

Hence $u = 3.3846 - 3.3 = 0.0846$

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