    Question

# A plane has a take-off speed of 120 ms−1 after covering a distance of 2400 m. Determine the acceleration (a) of the plane (assuming it to be constant) and the time (t) required to reach that speed.

A
a=4ms2,t=30 s
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B
a=3ms2,t=30 s
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C
a=4ms2,t=40 s
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D
a=3ms2,t=40 s
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Solution

## The correct option is D a=3ms−2,t=40 sGiven: Initial velocity, u=0 ms−1 Final velocity, v=120 ms−1 Distance covered, s=2400 m Let 'a' be the acceleration Let 't' be the time taken By the third equation of motion: v2=u2+2as 1202=02+2×a×2400 14400=4800a a=3 ms−2 By the first equation of motion: v=u+at 120=0+3t t=40 s  Suggest Corrections  0      Similar questions
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