Question

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

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

Answer 1

Answer: (C), (D)

The correct options are
C Average velocity is $\frac{5}{3}\text{m/s}$
D Acceleration is $4{\text{m/s}}^2$ at t = 0

Average velocity $\overrightarrow{v}=\frac{{\int }_0^{5}vdt}{{\int }_0^{5}dt}=\frac{{\int }_0^5(4t-t^2)dt}{{\int }_0^{5}dt}$
$=\frac{{[2t^2-\frac{t^3}{3}]}_0^5}{5}=\frac{50-\frac{125}{3}}{5}=\frac{25}{3\times 5}=\frac{5}{3}$
For average speed,
let us put $v=0$, which gives $t=0$ and $t=4s$.
$\therefore \text{Average speed}=\frac{\left|{\int }_0^{4}vdt\right|+\left|{\int }_4^{5}vdt\right|}{{\int }_0^{5}dt}$
$=\frac{\left|{\int }_0^4\right(4t-t^2\left)dt\right|+\left|{\int }_4^{5}vdt\right|}{5}$
$=\frac{{[2t^2-\frac{t^3}{3}]}_0^4+{[\frac{t^3}{3}-2t^2]}_4^5}{5}$
$=\frac{(32-\frac{64}{3})+\left[\frac{1}{3}\right(125-64)-2(25-16\left)\right]}{5}=\frac{13}{5}\text{m/s}$
For acceleration : $a=\frac{dv}{dt}=\frac{d}{dt}(4t-t^2)=4-2t$
At $t=0,a=4{\text{m/s}}^2$