Question

Consider a car initially at rest, starts to move along a straight road first with acceleration $5\, m / s ^{2}$, then with uniform velocity and finally, decelerating at $5\, m / s ^{2}$, before coming to a stop. Total time taken from start to end is $t=25\, s$. If the average velocity during that time is $72\, km / hr$, 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, $a=5\, m / s ^{2}$
deceleration of car $a=5\, m / s ^{2}$, total time taken from start of end is, $t=25\, s$ and average velocity of car,
$v _{ avg }=72\, km / hr =20\, m / s \,\left(\because l \frac{ km }{ hr }=\frac{5}{18} m / s \right)$
Since, $v _{\text {avg }}=\frac{\text { total displacement }}{\text { total time taken }}$
$2t$, total time taken by the car durig acceleration and deceleration
$v _{\text {avg }} =20=\frac{d_{t}+d_{(25-2 t)}+d_{t}}{25}$
$=\frac{2 d_{t}+d_{(25-2 t)}}{25}$
Since, $d_{t}=0+\frac{1}{2} a t^{2}=\frac{1}{2} a t^{2}=\frac{5}{2} t^{2}$ and
$d_{(25-2 t)}=v_{\text {uni. }}(25-2 t)$
where, $v_{\text {uni. }}=5t$
now, $v_{\text {avg }}=20=\frac{2\left(\frac{5}{2} t^{2}\right)+5 t(25-2 t)}{25}$
$\therefore 20 \times 25=5 t^{2}+5 t(25-2 t)$
$\Rightarrow 500=5 t^{2}+125 t-10 t^{2}$
$\Rightarrow t^{2}-25 t+100=0$
So, it gives $t=20$ and $5\, s$.
Hence, the time of uniform motion,
$t_{20}=25-2 t=25-2 \times 20=-15 s
$
( $\because$ Not possible )
$
or t_{5}=25-10=15 s
$