For motion of an object along \(x\)-axis, the velocity \(v\) depends on the displacement \(x\) as \(v=3…

For motion of an object along \(x\)-axis, the velocity \(v\) depends on the displacement \(x\) as \(v=3 x^{2}-2 x\). What is the acceleration at \(x=2 \mathrm{~m}\).What is the value of acceleration/10 ?

Solution

Given $v=3 x^{2}-2 x$, differentiating $v$, we get $\begin{aligned} \frac{d v}{d t}&=(6 x-2) \frac{d x}{d t}=(6 x-2) v \\ \Rightarrow a&=(6 x-2)\left(3 x^{2}-2 x\right) . \text{ Now put } x=2 \mathrm{~m} \\ \Rightarrow a&=(6 \times 2-2)\left(3(2)^{2}-2 \times 2\right)=80 \mathrm{~m} / \mathrm{s}^{2} \end{aligned}$ $\begin{aligned} \frac{a}{10} & = \frac{80}{10} \\ & =8 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}$

Asked in: JEE Mains - Motion In One Dimension - Chapter Test

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