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Question

Physics Question on Motion in a straight line

A car moves from XX to YY with a uniform speed vuv_{u} and returns to YY with a uniform speed vdv_{d}. The average speed for this round trip is

A

2vdvuvd+vu\frac{2 v_{d} v_{u}}{v_{d}+v_{u}}

B

νuvd\sqrt{\nu_{u} v_{d}}

C

vdvuvd+vu\frac{v_{d} v_{u}}{v_{d}+v_{u}}

D

vu+vd2\frac{v_{u}+v_{d}}{2}

Answer

2vdvuvd+vu\frac{2 v_{d} v_{u}}{v_{d}+v_{u}}

Explanation

Solution

Let t1t_{1} and t2t_{2} be times taken by the car to go from XX to YY and then from YY to XX respectively. Then, t1+t2=XYvu+XYvd=XY(vu+vdvuvd)t_{1}+t_{2}=\frac{X Y}{v_{u}}+\frac{X Y}{v_{d}}=X Y \cdot\left(\frac{v_{u}+v_{d}}{v_{u} v_{d}}\right)
Total distance travelled
=XY+XY=2XY=X Y+X Y=2 X Y
Therefore, average speed of the car for this round trip is
vav=2XYXY(vu+vdvuvd)v_{ av } =\frac{2 X Y}{X Y\left(\frac{v_{u}+v_{d}}{v_{u} v_{d}}\right)}
or vav=2vuvdvu+vdv_{ av }=\frac{2 v_{u} v_{d}}{v_{u}+v_{d}}