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Derivative from First Principle

The distance ...

Question

The distance covered by a particle in $t$ second is given by $x = 3 + 8 t - 4 t^{2}$ . After $1 s$ its velocity will be

0 u n i t / s

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3 u n i t / s

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4 u n i t / s

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7 u n i t / s

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Solution

The correct option is A $0 u n i t / s$
Given, $x = 3 + 8 t - 4 t^{2}$
On differentiating with respect to $t$ , we get
$v = \frac{d x}{d t} = 8 - 8 t$
At $t = 1, v = 8 - 8 = 0$ unit.

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