Question

A particle moves along a straight line and its velocity depends on time as $v=4t-t^2.$ Then for first $5\text{s}$

• A. Average velocity is $\frac{25}{3}\text{m}$
• B. Average speed is $10{\text{ms}}^{-1}$
• C. Average velocity is $\frac{5}{3}{\text{ms}}^{-1}$
• D. Acceleration is $4{\text{ms}}^{-2}\text{at}t=0$

Answer 1

Answer: (C), (D)

Acceleration is $4{\text{ms}}^{-2}\text{at}t=0$

Average velocity $\overrightarrow{v}=\frac{{\int }_0^{5}vdt}{{\int }_0^{5}dt}=\frac{{\int }_0^5(4t-t^2)dt}{{\int }_0^{5}dt}$
$=\frac{{[2t^2-\frac{t^3}{3}]}_0^5}{5}=\frac{50-\frac{125}{3}}{5}=\frac{25}{3\times 5}=\frac{5}{3}{\text{ms}}^{-1}$

For average speed,
let us put $v=0$, which gives $t=0$ and $t=4$ ie velocity changes sign at $t=4$.
$\therefore$ Average speed = $\frac{\left|{\int }_0^{4}vdt\right|+\left|{\int }_4^{5}vdt\right|}{{\int }_0^{5}dt}$
$=\frac{\left|{\int }_0^4\right(4t-t^2\left)dt\right|+\left|{\int }_4^{5}vdt\right|}{5}$
$=\frac{{[2t^2-\frac{t^3}{3}]}_0^4+{[-2t^2+\frac{t^3}{3}]}_4^5}{5}$
$=\frac{(32-\frac{64}{3})+[-2(25-16)+\frac{1}{3}(125-64\left)\right]}{5}=\frac{13}{5}{\text{ms}}^{-1}$

For acceleration:
$a=\frac{dv}{dt}=\frac{d}{dt}(4t-t^2)=4-2t$
At $t=0,a=4{\text{ms}}^{-2}$
$\therefore$ Options c and d are correct