Question

A body starts from rest and moves with a uniform acceleration of $3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.What will be its velocity at the end of $15seconds$?

Answer 1

Step 1: Given data

$Initialvelocity\left(u\right)=0m/s.$

$Acceleration,\left(a\right)=3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$

$Time,\left(t\right)=15s$

Step 2: Formula used

$v=u+at$ [ where V is the final velocity, u is the initial velocity, a is the acceleration and t is the time]

Step 3: Final Solution

$$\begin{gathered} v=u+at \\ v=0+3\times 15 \\ v=45\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Hence the velocity of the body is $45m/s.$