Question

A scooter is traveling at speed of $12\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$. If it is accelerating at $3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$, what is its velocity after$10seconds$?

Answer 1

Step 1:- Given data,

Initial velocity of scooter (u) =$12\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

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

Time = $10seconds.$

Step 2:- Formula,

$$\begin{gathered} Acceleration=\frac{(finalvelocity-initialvelocity)}{time} \\ a=\frac{(v-u)}{t} \end{gathered}$$

Step 3:- To find final velocity (v)

$$\begin{gathered} \Rightarrow a=\frac{(v-u)}{t} \\ \Rightarrow 3=\frac{(v-12)}{10} \\ \Rightarrow 3\times 10=(v-12) \\ \Rightarrow 30=(v-12) \\ \Rightarrow 30+12=v \\ \Rightarrow v=42\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Hence, final velocity(v) is $42\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.