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 $\frac{25}{3}m/s$
• B. Average speed is 10 m/s
• C. Average speed is $\frac{13}{5}m/s$
• D. Acceleration is $4{\text{m/s}}^2$ at t = 0

Answer 1

Answer: (C), (D)

The correct options are
C Average speed is $\frac{13}{5}m/s$
D Acceleration is $4{\text{m/s}}^2$ at t = 0

According to question
$v=4t-t^2$
Displacement $=dx=\int vdt$
Average velocity $\overrightarrow{v}=\frac{\text{Total displacement}}{\text{Total time}}=\frac{\underset{0}{\overset{5}{\int }}vdt}{\underset{0}{\overset{5}{\int }}dt}=\frac{\underset{0}{\overset{5}{\int }}(4t-t^2)dt}{\underset{0}{\overset{5}{\int }}dt}$
$=\frac{{[2t^2-\frac{t^3}{3}]}_0^5}{5}=\frac{50-\frac{125}{3}}{5}=\frac{25}{3\times 5}=\frac{5}{3}{\text{ms}}^{-1}$
Let us put $v=0$, we get
$4t-t^2=0$
$t=0$ and $t=4\text{s}$
After $t=4\text{s}$, the direction of motion will get reversed
$\therefore$ Average speed $=\frac{\left|\underset{0}{\overset{4}{\int }}vdt\right|+\left|\underset{4}{\overset{5}{\int }}vdt\right|}{\underset{0}{\overset{5}{\int }}dt}$
$=\frac{\left|\underset{0}{\overset{4}{\int }}(4t-t^2)dt\right|+\left|\underset{4}{\overset{5}{\int }}(4t-t^2)dt\right|}{5}$
$=\frac{{[2t^2-\frac{t^3}{3}]}_0^4+{[2t^2-\frac{t^3}{3}]}_4^5}{5}$
$=\frac{(32-\frac{64}{3})+\left|2\right(25-16)-\frac{1}{3}(125-64\left)\right|}{5}=\frac{13}{5}{\text{ms}}^{-1}$
For acceleration $a=\frac{dv}{dt}=\frac{d}{dt}(4t-t^2)=4-2t$
At $t=0,a=4{\text{m/s}}^2$
$\therefore$ option c and d are correct.