Problem

Problem

Velocity of a particle in xx - yy plane at any time tt is v=(2ti^+3t2j^) m/s\mathbf{v} = (2t\hat{i} + 3t^2\hat{j}) \text{ m/s} At t=0t = 0 , particle starts from the co-ordinates (2 m,4 m)(2 \text{ m}, 4 \text{ m}) . Find (a) acceleration of the particle at t=1 st = 1 \text{ s} . (b) position vector and co-ordinates of the particle at t=2 st = 2 \text{ s} .
Intuitive explanation

The key is recognizing that velocity is the derivative of position, so you need to reverse the process by integrating each component separately. Since you're given v(t)\mathbf{v}(t) and an initial position, integrate to find r(t)\mathbf{r}(t), then use the initial condition to find the integration constants. This transforms a calculus problem into finding position as a function of time.

Intuitive explanation

The key is recognizing that velocity is the derivative of position, so you need to reverse the process by integrating each component separately. Since you're given v(t)\mathbf{v}(t) and an initial position, integrate to find r(t)\mathbf{r}(t), then use the initial condition to find the integration constants. This transforms a calculus problem into finding position as a function of time.

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