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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 and the time (t) required to reach that speed.

A
a=4 ms2,t=30 s
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B
a=3 ms2,t=35 s
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C
a=3.5 ms2,t=40 s
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D
a=3 ms2,t=40 s
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Solution

The correct option is D a=3 ms2,t=40 s
Given:
Initial velocity, u=0 ms1
Final velocity, v=120 ms1
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 ms2

By the first equation of motion:
v=u+at
120=0+3t
t=40 s

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