If the speed of the particle is $v = 4 t^{3} - 6 t^{2} + 5 m/s$ , the distance of the particle as a function of time $t$ is

\frac{4 t^{4}}{3} - 3 t^{3} + 5 t m

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t^{4} - 2 t^{3} + 5 t m

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4 t^{4} + 6 t^{3} - 5 t m

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t^{4} - 3 t^{3} + 5 t m

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Solution

The correct option is B $t^{4} - 2 t^{3} + 5 t m$
Given, the speed of the car is $v (t) = 4 t^{3} - 6 t^{2} + 5$
So, the distance as a function of time is given as
$\int v (t) d t$
$\Rightarrow S = \int (4 t^{3} - 6 t^{2} + 5) d t = t^{4} - 2 t^{3} + 5 t m$

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