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(C) Let the initial velocity be $\vec{v}_1 = 50 \; 	ext{km/h}$ due north.After turning left by $90^{\circ}$,the final velocity is $\vec{v}_2 = 50 \;

Vedclass · Accepted Answer

$70.7 \; km/h$ along south-west direction

Vedclass · Answer

(C) Let the initial velocity be $\vec{v}_1 = 50 \; 	ext{km/h}$ due north.After turning left by $90^{\circ}$,the final velocity is $\vec{v}_2 = 50 \; 	ext{km/h}$ due west.The change in velocity is given by $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1 = \vec{v}_2 + (-\vec{v}_1)$.Here,$-\vec{v}_1 = 50 \; 	ext{km/h}$ due south.The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{v_2^2 + v_1^2} = \sqrt{50^2 + 50^2} = \sqrt{2500 + 2500} = \sqrt{5000} \approx 70.7 \; 	ext{km/h}$.The direction of $\Delta \vec{v}$ is the resultant of a vector pointing west and a vector pointing south,which is in the south-west direction.

$A$ bus is moving on a straight road towards north with a uniform speed of $50 \; km/h$. Then it turns left through $90^{\circ}$. If the speed remains unchanged after turning,the increase in the velocity of the bus in the turning process is:

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