Question

A body with an initial velocity of $3m{s}^{-1}$ moves with an acceleration of $2m{s}^{-2}$ then the distance travelled in the fourth second is what?

Answer 1

Step1: Given

Initial speed , $u=3m{s}^{-1}$

Acceleration, $a=2m{s}^{-2}$

time, $t=4^{th}s$

Step 2: Formula used and calculation

$S_n=u+\frac{1}{2}a(t-1)$

where $S_n$ =distance travelled

u=initial velocity

a=acceleration

t=time

Now put the value in the formula:

$S_n=3+\frac{1}{2}\times 2(4-1)$

$S_n=3+3$

$S_n=6m$

So, distance travelled in fourth second is

$6m$

.