Question

A particle moves along a straight line and its velocity depends on time ${}^{\prime }{t}^{\prime }$ as $v$ = $4t-t^2.$ Here $v$ is in $m/sec$ and $t$ is in second. Then for the first 5 seconds :

• A. Magnitude of average velocity is $\frac{5}{3}$ m/s
• B. Average speed is $\frac{13}{5}$ m/s
• C. Average speed is $\frac{11}{5}$ m/s
• D. Average acceleration is $-1m/s^2$

Answer 1

Answer: (A), (B), (D)

The correct options are
A Magnitude of average velocity is $\frac{5}{3}$ m/s
B Average speed is $\frac{13}{5}$ m/s
D Average acceleration is $-1m/s^2$

Average velocity $=\frac{S}{\Delta t}=V_{avg}$

Displacement, $S={\int }_0^{5}vdt={\int }_0^5(4t-t^2)dt=\frac{25}{3}m$
$V_{avg}=\frac{25/3m}{5sec.}=\frac{5}{3}\frac{m}{sec}$
Now,
$v=0att=4s$
And after $4s$ the direction of velocity changes
$\therefore$Average speed $=\frac{distancecovered}{timetaken}=\frac{distance}{\Delta t}$
Distance $={\int }_0^{4}vdt+{\int }_4^5(-v)dt$
$=\frac{32}{3}+\frac{7}{3}=\frac{39}{3}m=13m$
Average speed $=\frac{13}{5}m/s$
Average acceleration $\left(a_{avg}\right)=\frac{V_f-V_i}{\Delta t}$
$V_f=4\times 5-5^2=20-25=-5$
$V_1=0$
$A_{avg}=\frac{-5-0}{5}=-1m/s^2$.