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21 April, 22:27

Determine the displacement of a plane that is uniformly accelerated from 66 m/s to 88 m/s in 12 s

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  1. 21 April, 22:43
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    Displacement is the area under the graph for velocity-time graph so if plane accelerated uniformly from 66m/s to 88m/s the area will be the area of trapezium where formula for area under trapezium is 1/2 (sum of parallel lines) x height so the parallel lines in this case are 66m/s and 88m/s and height is time taken which is 12 so displacement is 1/2 (66+88) (12) = 924m

    if you want to know the formula then it is S=1/2 (V+U) (t) where S is displacement V is final velocity U is initial velocity and t is time taken
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