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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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Respuesta :

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
okpalawalter8

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

d=263.52m

From the question we are told

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

Generally the equation for the  acceleration is mathematically given as[tex]a=\frac{u-v}{t}\\\\\a=\frac{88-66}{12}\\\\a=1.83m/s^2[/tex]

Therefore

[tex]a=\frac{d}{t^2}\\\\\d=1.83*12^2\\\\\d=263.52m[/tex]

Therefore

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

d=263.52m

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