The position $x$ of a particle varies with time $(t)$ as $x=a t^{2}-b t^{3} .$ The acceleration at time $t$ of the particle will be equal to zero, where $t$ is equal to

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Aicharya · Accepted Answer

$\begin{array}{l} x=a t^{2}-b t^{3} \ \text {velocity }=\frac{d x}{d t}=2 a t-3 b t^{2} \ \text { and acceleration }=\frac{d}{d t}\left(\frac{d^{2} x}{d t^{2}}\right)=2 a-6 b t \end{array}$ Acceleration will be zero if $2 a-6 b t=0 \Rightarrow t=\frac{2 a}{6 b}=\frac{a}{3 b}$

The position \(x\) of a particle varies with time \((t)\) as \(x=a t^{2}-b t^{3} .\) The acceleration at…

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