Question

Consider a car initially at rest, starts to move along a straight road first with acceleration $5m/s^2$, then with uniform velocity and finally, decelerating at $5m/s^2$, before coming to a stop. Total time taken from start to end is $t=25s$. If the average velocity during that time is $72km/hr$, the car moved with uniform velocity for a time of

• A. 15 s
• B. 30 s
• C. 155 s
• D. 2 s

Answer 1

Answer: (A)

Given, Acceleration of car, $a=5m/s^2$
deceleration of car $a=5m/s^2$, total time taken from start of end is, $t=25s$ and average velocity of car,
$v_{avg}=72km/hr=20m/s\left(\because l\frac{km}{hr}=\frac{5}{18}m/s\right)$
Since, $v_{\text{avg }}=\frac{\text{ total displacement }}{\text{ total time taken }}$
$2t$, total time taken by the car durig acceleration and deceleration
$v_{\text{avg }}=20=\frac{d_t+d_{(25-2t)}+d_t}{25}$
$=\frac{2d_t+d_{(25-2t)}}{25}$
Since, $d_t=0+\frac{1}{2}a{t}^2=\frac{1}{2}a{t}^2=\frac{5}{2}{t}^2$ and
$d_{(25-2t)}=v_{\text{uni. }}(25-2t)$
where, $v_{\text{uni. }}=5t$
now, $v_{\text{avg }}=20=\frac{2\left(\frac{5}{2}{t}^2\right)+5t(25-2t)}{25}$
$\therefore 20\times 25=5t^2+5t(25-2t)$
$\Rightarrow 500=5t^2+125t-10t^2$
$\Rightarrow t^2-25t+100=0$
So, it gives $t=20$ and $5s$.
Hence, the time of uniform motion,
$t_{20}=25-2t=25-2\times 20=-15s<br/>$
( $\because$ Not possible )
$<br/>or{t}_5=25-10=15s<br/>$