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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?`

A

`2 ms ^(-1), 5ms^(-1)`

B

`5 ms ^(-1), 10ms^(-1)`

C

`10 ms ^(-1), 15ms^(-1)`

D

`10 ms ^(-1), 20ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
Here , `x = a+b^(2) , `
where `a = 8.5 m and b = 2.5 ms^(-2)`
Velocity, `v=(dx)/(dt)=d/(dt) (a+bt^2)=2bt=5t ms^(-1)`
At `t = 2 s, v = 5 (2) ms^(-1) = 10 ms ^(-1) `
Average velocity `=(x(4)-x(2))/(4-2)=((a+16b)-(a-4b))/2=6b`
`= 6(2.5)ms ^(-1) = 15 ms^(-1)`
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