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The position of an object moving along x...

The position of an object moving along x-axis is given by `x=a+bt^(2)` where `a = 8.5 m, b = 2.5 m s^(-2)` and t is measured in seconds. What is its velocity at t = 0 s and t = 2.0 s. What is the average velocity between t = 2.0 s and t = 4.0 s ?

Text Solution

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In notation of differential calculus, the velocity is
`upsilon = (dx)/(dt)=(d)/(dt)(a+bt^(2))=2b t=5.0 t ms^(-1)`
At t = 0 s, `" " upsilon = 0 ms^(-1)` and at t = 2.0 s, `upsilon =10 ms^(-1)`.
Average velocity `=(x(4.0)-x(2.0))/(4.0-2.0)`
`=(a+16b-a-4b)/(2.0)=6.0xx b`
`=6.0xx2.5=15 ms^(-1)`.
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