Discussion Forum

Consider a car initially at rest, starts to move along a straight road first with acceleration 5 m/s² , then with uniform velocity and finally, decelerating at 5 m/s², 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 

Answer:

Explanation:

 Given , acceleration of car, a= 5 m/s²,

 deceleration of   car a=5 m/s² , total time taken from start of end is, t=25s and average velocity of car

   $v_{avg}$=72 km/hr=20 m/s   ( 1$\frac{km}{hr}=\frac{5}{18}m/s$)

Since, $v_{avg}$= total displacement/total time taken

 2t, total time taken by the car during acceleration  and deceleration,

 $v_{avg}=20=\frac{d_{1}+d_{(25-2t)}+d_{t}}{25}$

     $=\frac{2d_{1}+d_{(25-2t)}}{25}$

 Since, $d_{t}=0+\frac{1}{2}{at^{2}}=\frac{1}{2}at^{2}=\frac{5}{2}t^{2} $   and  

    $d_{(25-2t)}= v_{uni.}(25-2t)$

where, $v_{uni.} $=5t

now,   $v_{avg}=20=\frac{2\left(\frac{5}{2}t^{2}\right)+5t(25-2t)}{25}$

$\therefore$       $20 \times 25 =5t^{2}+5t(25-2t)$

$\Rightarrow$  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,

 $t_{20}=25-2t=25-2 \times 20=-15  s$

                                   ($\because$ Not possible )

 or $t_{5}=25-10=15s$

So, the correct option is (a)