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Please solve RD Sharma class 12 chapter 12 Derivative as a rate measure exercise Very short answer type question  1 maths textbook solution

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Answer: 9 Units/unit time.

Hint: Here we use the concept of rate of change of bodies and quantities

Given:  s=t^{3}-6 t^{2}+9 t+8

Solution:  Here,

        s=t^{3}-6 t^{2}+9 t+8

Let’s differentiate along time(t)

        \frac{d s}{d t}=3 t^{2}-12 t+9

Initial velocity= velocity at (t=0)

        \left.\frac{d s}{d t} \text { (at } \mathrm{t}=0\right)=9 \text { units/unit time }

So, the initial velocity of the particle is 9 units/unit time.


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