A particle is moving in one dimension (along x axis) under the ac… - Vidyadip

MCQ

A particle is moving in one dimension (along $\mathrm{x}$ axis) under the action of a variable force. It's initial position was $16 \mathrm{~m}$ right of origin. The variation of its position ( $\mathrm{x}$ ) with time ( $\mathrm{t})$ is given as $\mathrm{x}=-3 \mathrm{t}^5+18 \mathrm{t}^2+16 \mathrm{t}$, where $\mathrm{x}$ is in $\mathrm{m}$ and $\mathrm{t}$ is in $\mathrm{s}$. The velocity of the particle when its acceleration becomes zero is________ $\mathrm{m} / \mathrm{s}.$

A
$50$
✓
$52$
C
$57$
D
$60$

✓

Answer

Correct option: B.

$52$

b
$x=3 t^3+18 t^2+16 t$

$v=-9 \mathrm{t}^2+36+16$

$a=-18 \mathrm{t}+36$

$a=0 \text { at } \mathrm{t}=2 \mathrm{~s}$

$v=-9(2)^2+36 \times 2+16$

$v=52 \mathrm{~m} / \mathrm{s}$

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