Question

a man runs at a speed of 4m/s to a board a standing bus. when he is 6 m behind the door at t=0, the bus moves forward and continues with acceleration of 1.2 m/s^2 . what will be the distance covered

Answer 1

At t=0 man's position is origin and that of the bus is 6 m. The equations of motions are:

xm=xmo+vmot+12amt2xm=xmo+vmot+12amt2

xb=xbo+vbot+12abt2xb=xbo+vbot+12abt2

vmo=4m/s;vbo=0vmo=4m/s;vbo=0

am=0;ab=1.2m/s2am=0;ab=1.2m/s2

xm=4.0txm=4.0t

xb=6.0+0.6t2xb=6.0+0.6t2

xm=xb⇒4.0t=6.0+0.6t2xm=xb⇒4.0t=6.0+0.6t2

t=20±400−360√6=10±10√3=2.3sand4.3st=20±400−3606=10±103=2.3sand4.3s

At 2.3 s the velocity of the bus is greater tham that of the man so man cannot catch the bus. But after 4.3 s the relative velocity of the bus and man is zero so catching of the bus is possible. Soo correcct answer is (b) 4.3 s