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Provide solution for RD Sharma maths class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 18

Answers (1)

Answer:

D. \: \: -\frac{16}{3}\: unit/sec

Hint:

We are given the distance travelled as a function of time, We can calculate velocity and acceleration by

\begin{aligned} &V t=\frac{d s}{d t} \text { and } a=\frac{d^{2} s}{d t^{2}} \\ \end{aligned}

Given:

\begin{aligned} s t=t^{3}-4 t^{2}+5 \end{aligned}

Solution:

→ Differentiating with respect to time, we get

\begin{aligned} &V t=\frac{d s}{d t}=3 t^{2}-8 t \end{aligned}

→ Differentiating again with respect to time, we get

\begin{aligned} &a(t)=\frac{d^{2} s}{d t^{2}}=6 t=8 \end{aligned}

\begin{aligned} &\rightarrow \text { Given that } a=0 \Rightarrow 6 t-8=0 \\ &\text { Or }=\frac{4}{3} \text { unit } / \mathrm{sec} \\ &V\left(\frac{4}{3}\right)=3 \times \frac{4^{2}}{3^{2}}-8 \times \frac{4}{3}=-\frac{16}{3} \text { unit } / \mathrm{sec} \end{aligned}

Posted by

Gurleen Kaur

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