A bus starts from rest and accelerates at a uniform rate 4 m/s2 for certain time. It then moves with a constant speed for some time and finally retards at 4 m/s2 to come to rest. Average speed of the bus during the total journey is l5 m/s and the total time is 20 sec. Time duration for which the car moves with constant speed is:

vot1=4⇒vo = 4t1Average veloclty=DisplacementTime taken15=12vot1+vot+12vot1t1+t+t1=4t1t1+t202t1+t=20⇒t1=20−t2=10−t2Therefore, 300=410−t210−t2+t⇒75=10−t210−t2+t=100−t24t2 = 100t = 10s

A bus starts from rest and accelerates at a uniform rate 4 m/s2 for certain time. It then moves with a constant speed for some time and finally retards at 4 m/s2 to come to rest. Average speed of the bus during the total journey is l5 m/s and the total time is 20 sec. Time duration for which the car moves with constant speed is:

vot1=4⇒vo = 4t1Average veloclty=DisplacementTime taken15=12vot1+vot+12vot1t1+t+t1=4t1t1+t202t1+t=20⇒t1=20−t2=10−t2Therefore, 300=410−t210−t2+t⇒75=10−t210−t2+t=100−t24t2 = 100t = 10s

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A bus starts from rest and accelerates at a uniform rate 4 m/s²for certain time. It then moves with a constant speed for some time and finally retards at 4 m/s² to come to rest. Average speed of the bus during the total journey is l5 m/s and the total time is 20 sec. Time duration for which the car moves with constant speed is:

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answer is D.

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Detailed Solution

$\frac{v_{o}}{t_{1}} = 4$
$\Rightarrow v_{o} = 4 t_{1}$
$Average veloclty = \frac{Displacement}{Time taken}$
$15 = \frac{\frac{1}{2} v_{o} t_{1} + v_{o} t + \frac{1}{2} v_{o} t_{1}}{t_{1} + t + t_{1}}$
$= \frac{4 t_{1} (t_{1} + t)}{20}$
$2 t_{1} + t = 20$
$\Rightarrow t_{1} = \frac{20 - t}{2} = 10 - \frac{t}{2}$
$Therefore, 300=4 (10 - \frac{t}{2}) (10 - \frac{t}{2} + t)$
$\Rightarrow 75 = (10 - \frac{t}{2}) (10 - \frac{t}{2} + t) = 100 - \frac{t^{2}}{4}$
t² = 100
t = 10s