Question

The position $x$ of a moving particle at time $t$ is given by $x=t^2+2t+3\text{m}$. The average velocity of the particle during the first second of its motion is

• A. $1{\text{ms}}^{-1}$
• B. $2{\text{ms}}^{-1}$
• C. $3{\text{ms}}^{-1}$
• D. $4{\text{ms}}^{-1}$

Answer 1

The correct option is C $3{\text{ms}}^{-1}$

Given, $x=t^2+2t+3\text{m}$
$V_{\text{avg}}=\frac{\text{Displacement}}{\text{Time}}$
Displacement = Change in position = $x_1-x_0$
At $t=0s$, $x_0=3\text{m}$
and at $t=1sec$ , $x_1=1+2+3=6\text{m}$
$V_{\text{av}}=\frac{6-3}{1}=3{\text{ms}}^{-1}$