A car is moving with a velocity of 4 ,m ,s^{- | vedclass

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Question

$(A)$ The initial velocity vector is $\vec{v}_i = 4 \hat{i} \,m \,s^{-1}$.The final velocity vector is $\vec{v}_f = 4 \sqrt{2} (\cos 45^\circ \hat{i}

Vedclass · Accepted Answer

$2 \sqrt{5} \,m \,s^{-1}$

Vedclass · Answer

$(A)$ The initial velocity vector is $\vec{v}_i = 4 \hat{i} \,m \,s^{-1}$.The final velocity vector is $\vec{v}_f = 4 \sqrt{2} (\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) = 4 \sqrt{2} (\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}) = 4 \hat{i} + 4 \hat{j} \,m \,s^{-1}$.The displacement $\vec{s}$ is the integral of velocity over time. Assuming constant acceleration, $\vec{s} = \vec{v}_{avg} 	imes t = \frac{\vec{v}_i + \vec{v}_f}{2} 	imes t$.$\vec{v}_{avg} = \frac{(4 \hat{i}) + (4 \hat{i} + 4 \hat{j})}{2} = \frac{8 \hat{i} + 4 \hat{j}}{2} = 4 \hat{i} + 2 \hat{j} \,m \,s^{-1}$.The magnitude of the average velocity is $|\vec{v}_{avg}| = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2 \sqrt{5} \,m \,s^{-1}$.

$A$ car is moving with a velocity of $4 \,m \,s^{-1}$ towards east. After a time of $4 \,s$, if it is heading north-east with a velocity of $4 \sqrt{2} \,m \,s^{-1}$, then the average velocity of the car is

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