A plane has a take-off speed of 120ms−1 after covering a distance of 2400m. Determine the acceleration (a) of the plane and the time (t) required to reach that speed.
A
a=4ms−2,t=30s
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B
a=3ms−2,t=35s
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C
a=3.5ms−2,t=40s
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D
a=3ms−2,t=40s
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Solution
The correct option is Da=3ms−2,t=40s Given: Initial velocity, u=0ms−1 Final velocity, v=120ms−1 Distance covered, s=2400m
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=3ms−2
By the first equation of motion: v=u+at 120=0+3t t=40s