Question

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

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

Answer 1

Answer: (A), (B), (C)

Average velocity $=\frac{s}{\Delta t}=v_{avg}$
$S={\int }_0^{5}vdt={\int }_0^5(4t-t^2)dt=\frac{25}{3}m$
$v_{avg}=\frac{25/3m}{5sec}=\frac{5}{3}\frac{m}{sec}$
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 $=\frac{13m}{5sec}$
Average acceleration $\left(a_{avg}\right)=\frac{v_f-v_i}{\Delta t}$
$v_f=4\times 5-5^2=20-25=-5$
$v_i=0$
$a_{avg}=\frac{-5-0}{5}=-1m/s^2$