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Physics Question on Motion in a straight line

A jet airplane travelling at the speed of 500 km h−1500 \,km \,h^{-1} ejects its products of combustion at the speed of 1500 km h−11500 \,km\, h^{-1} relative to the jet plane. The speed of the products of combustion with respect to an observer on the ground is

A

500 km h−1500 \,km\, h^{-1}

B

1000 km h−11000 \,km\, h^{-1}

C

1500 km h−11500 \,km\, h^{-1}

D

2000 km h−12000 \,km\, h^{-1}

Answer

1000 km h−11000 \,km\, h^{-1}

Explanation

Solution

Velocity of jet plane w.r.t. ground vJG=500 km h−1v_{JG} = 500 \,km\, h^{-1} Velocity of products of combustion w.r.t. jet plane vCJ=−1500 km h−1v_{CJ} = - 1500\, km \,h^{-1} ∴\therefore Velocity of products of combustion w.r.t. ground is vCG=vCJ+vJG=−1500 km h−1+500 km h−1v_{CG}=v_{CJ}+v_{JG}=-1500\,km\,h^{-1}+500\,km\,h^{-1} =−1000 km h−1=-1000\,km\,h^{-1} −ve-ve sign shows that the direction of products of combustion is opposite to that of the plane. ∴\therefore Speed of the products of combustion w.r.t. ground =1000 km h−1= 1000\, km \,h^{-1} .