The acceleration of a particle varies with time $t$ seconds according to the relation $a = 6 t + 6 m s^{- 2}$ . Find velocity and position as functions of time. It is given that the particle starts from origin at $t = 0$ with velocity $2 m s^{- 1}$ .

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Solution

We know that $\int d v = \int a d t \Rightarrow \int_{2}^{v} d v = \int_{0}^{t} (6 t + 6) d t$
$\Rightarrow v - 2 = {[\frac{6 t^{2}}{2} + 6 t]}_{0}^{t} \Rightarrow v = 3 t^{2} + 6 t + 2$
Now $\int d x = \int v d t \Rightarrow \int_{0}^{x} d x = \int_{0}^{t} (3 t^{2} + 6 t + 2) d t$
$\Rightarrow x = t^{3} + 3 t^{2} + 2 t$

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The acceleration of a particle varies with time t seconds according to the relation a=6t+6 ms−2. Find velocity and position as functions of time. It is given that the particle starts from origin at t=0 with velocity 2 ms−1.

We know that ∫dv=∫adt⇒∫v2dv=∫t0(6t+6)dt⇒v−2=[6t22+6t]t0⇒v=3t2+6t+2Now ∫dx=∫vdt⇒∫x0dx=∫t0(3t2+6t+2)dt⇒x=t3+3t2+2t

The acceleration of a particle varies with time $t$ seconds according to the relation $a = 6 t + 6 m s^{- 2}$ . Find velocity and position as functions of time. It is given that the particle starts from origin at $t = 0$ with velocity $2 m s^{- 1}$ .