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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

velocity at t = 4 sec is 10 m/s

Velocity at t=2 sec is zero

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

answer is B.

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Detailed Solution

$\large v = \frac{{dx}}{{dt}} = 2bt{\text{ = 2 }} \times {\text{ 2}}{\text{.5 }} \times {\text{ t}} = 5t$

$\large t = 2 \Rightarrow {x_2} = 8.5 + 4 \times 2.5$

$\large t = 4 \Rightarrow {x_4} = 8.5 + 16 \times 2.5$

$\large {v_{avg}} = \frac{{{x_4} - {x_2}}}{{4 - 2}} = \frac{{\cancel{{8.5}} + 16 \times 2.5 - \cancel{{8.5}} - 4 \times 2.5}}{{\cancel{2}}}$

$\large = \frac{{12 \times 2.5}}{2} = 6 \times 2.5 \Rightarrow {v_{avg}} = 15{\kern 1pt} m/s$

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