Question

A particle's velocity changes from (2hat I +3 hatj) ms^(-1) in to (3 hati - 2hatj) ms^(-1) in 2s. Its average acceleration in ms^(-2) is

Answer 1

Given:

• Initial velocity, $\mathsf{\vec{v}_{1}=2\hat{i}+3\hat{j}\:ms^{-1}}$
• Terminal velocity, $\small{\mathsf{\vec{v}_{2}=3\hat{i}-2\hat{j}\:ms^{-1}}}$
• Time, $\mathsf{t=2\:s}$

To find: average acceleration in $\mathsf{ms^{-2}}$

Solution:

Here, average acceleration

$\quad\mathsf{=\dfrac{\vec{v}_{2}-\vec{v}_{1}}{t}\:ms^{-2}}$

$\quad\mathsf{=\dfrac{(3\hat{i}-2\hat{j})-(2\hat{i}+3\hat{j})}{2}\:ms^{-2}}$

$\quad\mathsf{=\dfrac{3\hat{i}-2\hat{j}-2\hat{i}-3\hat{j}}{2}\:ms^{-2}}$

$\quad\mathsf{=\dfrac{\hat{i}-5\hat{j}}{2}\:ms^{-2}}$