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Work Done as Dot Product

Displacement ...

Question

Displacement of a particle of mass $1 kg$ moving in a straight line varies with time $^{'} t^{'}$ as,
$s = [2 t^{3} + 4] m$ .
Impulse of the force acting on the particle over a time interval $t = 0 s and t = 1 s$ will be Ns (Write upto two digits after the decimal point.)

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Solution

We have $v = \frac{d s}{d t} = 6 t^{2}$
Impulse = $Δ p = m (Δ V) = 1 [6 - 0] = 6 Ns$

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