A jet airplane travelling from east to west at a speed of 500kmh−1 eject out gases of combustion at a speed of 1500kmh−1 with respect to the jet plane. What is the velocity of the gases with respect to an observer on the ground?
A
1000kmh−1 in the direction west to east.
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B
1000kmh−1 in the direction east to west.
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C
2000kmh−1 in the direction west to east.
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D
2000kmh−1 in the direction east to west.
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Solution
The correct option is A1000kmh−1 in the direction west to east. Suppose we choose the positive direction to be from east to west. Then the velocity of the plane =+500kmh−1. Since, the gases are ejected in the direction opposite to the direction of motion of the plane, the relative velocity of the gases with respect to the plane =−1500kmh−1. Therefore the velocity of the gases with respect to an observer on the ground =−1500+500=−1000kmh−1 The negative sign indicates that the direction of the velocity is from west to east. Hence, the correct option is A.