A particle is moving in a straight line. The variation of position as a function of time is given as m. The velocity of the body when its acceleration becomes zero is

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Solution

Q.
A particle is moving in a straight line. The variation of position $' x'$ as a function of time $' t'$ is given as $x = (t^{3} - 6 t^{2} + 20 t + 15)$ m. The velocity of the body when its acceleration becomes zero is [2024]

1 6 m/s

2 10 m/s

3 8 m/s

4 4 m/s

Ans.
(3)

    $x = t^{3} - 6 t^{2} + 20 t + 15$

$\frac{d x}{d t} = v = 3 t^{2} - 12 t + 20$

   $\frac{d v}{d t} = a = 6 t - 12$

When $a = 0$

   $t = 2 s e c$

At $t = 2 s e c$

$v = 3 {(2)}^{2} - 12 (2) + 20 = 8 m / s$

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