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The position of and object moving along x-axis is gi en by `x=a +bt^(2)`, wher `a=8.5 m` and b=2.5 ms^(-2) and (t) is measured in ceconds. What is the velcoity `at t=1= 0`s and `t=2.0 s?` What is the average velocity between `t=2.0s` and `t=4.0 s`?

Answer» In notation of differential calculus, the velocity is
`v = (dx)/(dt) = (d)/(dt) (a + bt^(2)) = 2b t = 5.0 ms^(-1)`
At `t = 0s , v = 0 m s^(-1)` and at `t = 2.0 s, v = 10 ms^(-1)`
Average velocity `= (x(4.0) - x (2.0))/(4.0 - 2.0)`
`= (a + 16 b - a 4 b)/(2.0) = 6.0 xx b`
`= 6.0 xx 2.5 = 15 ms^(-1)`


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