The position of an object moving along x-axis is given by x = a + bt² where a = 8.5 m, b = 2.5 ms^–2 and t is measured in seconds. Then which of the following is true ?

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all the above are true

average velocity between t = 2, t = 4 sec is 15 m/s.

velocity at t = 4 sec is 10 m/s

Velocity at t=2 sec is zero

answer is B.

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Detailed Solution

$\begin{array}{l} v = \frac{dx}{dt} = 2 bt = 2 \times 2.5 \times t = 5 t \\ t = 2 \Rightarrow x_{2} = 8.5 + 4 \times 2.5 \\ t = 4 \Rightarrow x_{4} = 8.5 + 16 \times 2.5 \\ v_{avg} = \frac{x_{4} - x_{2}}{4 - 2} = \frac{8.5 + 16 \times 2.5 - 8.5 - 4 \times 2.5}{22} \\ = \frac{12 \times 2.5}{2} = 6 \times 2.5 \Rightarrow v_{avg} = 15 m / s \end{array}$