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2.Motion in Straight Line
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The position of a particle moving along $x$-axis given by $x=\left(-2 t^3+3 t^2+5\right) m$. The acceleration of particle at the instant its velocity becomes zero is ....... $m / s ^2$
A$12$
B$-12$
C$-6$
D$0$
Solution
(c)
$x=\left(-2 t^3+3 t^2+5\right) m$
$\Rightarrow \frac{d x}{d t}=-6 t^2+6 t=v$
$\Rightarrow \frac{d^2 x}{d t^2}=-12 t+6 \quad\left(\text { for } v=0,6 t=6 t^2 \Rightarrow t=1 s \right)$
$\left.a\right|_{t=1 s }=-12+6=-6 \,ms ^{-2}$
$x=\left(-2 t^3+3 t^2+5\right) m$
$\Rightarrow \frac{d x}{d t}=-6 t^2+6 t=v$
$\Rightarrow \frac{d^2 x}{d t^2}=-12 t+6 \quad\left(\text { for } v=0,6 t=6 t^2 \Rightarrow t=1 s \right)$
$\left.a\right|_{t=1 s }=-12+6=-6 \,ms ^{-2}$
Standard 11
Physics
