Question

9. A particle moving in the x y plane$\overrightarrow{v}=\left(2t\hat{i}+\left(3-t^2\right)\hat{j}\right)\text{m}/\text{s}$

Here t is in s. What is the particle's acceleration vector at $t=4\text{s}$?

Short answer

The acceleration vector of the particle at is$(2\widehat{i}-8\widehat{j})\text{m}/{\text{s}}^2$

Step-by-step solution

Step 1. Given information

The velocity vector of the particle moving in the x-y plane is $\overrightarrow{v}=\left(2t\widehat{i}+\left(3-t^2\right)\widehat{j}\right)\text{m}/\text{s}$and the time instant is .

$t=4\text{s}$

Step 2. Explanation

The acceleration vector of the particle is,

$\overrightarrow{a}=\frac{d}{dt}\left(\overrightarrow{v}\right)$

$\overrightarrow{a}=\frac{d}{dt}\left(\left(2\widehat{i}+\left(3-t^2\right)\widehat{j}\right)\text{m}/\text{s}\right)$

For t = 4s

The acceleration vector of the particle is

$\begin{array}{l}\overrightarrow{a}=(2\widehat{i}-2(4\text{s}\left)\widehat{j}\right)\text{m}/{\text{s}}^2\\ =(2\widehat{i}-8\widehat{j})\text{m}/{\text{s}}^2\end{array}$

Therefore, the acceleration vector of the particle is$(2\widehat{i}-8\widehat{j})\text{m}/{\text{s}}^2$