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Physics Question on Motion in a straight line

A police van moving on a highway with a speed of 30 km h−130\, km\, h^{-1} fires a bullet at a thief?s car speeding away in the same direction with a speed of 192 km h−1192\, km\, h^{-1} . If the muzzle speed of the bullet is 150 m s−1150 \,m \,s^{-1} , with what speed does the bullet hit the thief?s car ?

A

95 m s−195 \,m \,s^{-1}

B

105 m s−1105 \,m \,s^{-1}

C

115 m s−1115 \,m \,s^{-1}

D

125 m s−1125 \,m \,s^{-1}

Answer

105 m s−1105 \,m \,s^{-1}

Explanation

Solution

Speed of police van w.r.t. ground ∴vPG=30 km h−1\therefore v_{PG}=30\,km\,h^{-1} Speed of thief?s car w.r.t. ground ∴vTG=192 km h−1\therefore v_{TG}=192\,km\,h^{-1} Speed of bullet w.r.t. police van vBP=150 ms−1=150×185km h−1=540 km h−1v_{BP}=150\,ms^{-1}=150\times\frac{18}{5}km\,h^{-1}=540\,km\,h^{-1} Speed with which the bullet will hit the thief?s car will be vBT=vBG+vGT=vBP+vPG+vGTv_{BT}=v_{BG}+v_{GT}=v_{BP}+v_{PG}+v_{GT} =540 km h−1+30 km h−1−192 km h−1=540\,km\,h^{-1}+30\,km\,h^{-1}-192\,km\,h^{-1} (∵vGT=−vTG)\left(\because v_{GT}=-v_{TG}\right) =378 km h−1=378×518 ms−1=105 ms−1=378\,km\,h^{-1}=378\times\frac{5}{18}\,ms^{-1}=105\,ms^{-1}