A car moves with a speed of 20 , m/s towards | vedclass

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(D) Let the initial velocity be $\vec{v}_1 = 20\hat{j} \, m/s$ (towards North).After turning,the final velocity is $\vec{v}_2 = -20\hat{i} \, m/s$ (to

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$20\sqrt{2} \, m/s \, S-W$

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(D) Let the initial velocity be $\vec{v}_1 = 20\hat{j} \, m/s$ (towards North).After turning,the final velocity is $\vec{v}_2 = -20\hat{i} \, m/s$ (towards West).The change in velocity is given by $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$.$\Delta \vec{v} = -20\hat{i} - 20\hat{j} = -20(\hat{i} + \hat{j}) \, m/s$.The magnitude of change in velocity is $|\Delta \vec{v}| = \sqrt{(-20)^2 + (-20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \, m/s$.The direction is given by $	an 	heta = \frac{|\Delta v_y|}{|\Delta v_x|} = \frac{20}{20} = 1$,so $	heta = 45^\circ$.Since both components are negative,the direction is South-West $(S-W)$.

$A$ car moves with a speed of $20 \, m/s$ towards the North. It turns and moves with the same speed towards the West. What is the change in velocity?

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