A scooter going due east at 10, ms^{-1} turns | vedclass

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Question

(D) Initial velocity $\vec{v_1} = 10\hat{i}\, ms^{-1}$ (east).Final velocity $\vec{v_2} = 10(-\hat{j}) = -10\hat{j}\, ms^{-1}$ (south).The change in v

Vedclass · Accepted Answer

$14.14\, ms^{-1}$ in south-west direction

Vedclass · Answer

(D) Initial velocity $\vec{v_1} = 10\hat{i}\, ms^{-1}$ (east).Final velocity $\vec{v_2} = 10(-\hat{j}) = -10\hat{j}\, ms^{-1}$ (south).The change in velocity is given by $\Delta \vec{v} = \vec{v_2} - \vec{v_1}$.$\Delta \vec{v} = -10\hat{j} - 10\hat{i} = -(10\hat{i} + 10\hat{j})$.The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{(-10)^2 + (-10)^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \approx 14.14\, ms^{-1}$.The direction is south-west because both components are negative.Alternatively,using the formula for change in velocity when speed $v$ is constant and angle is $	heta$: $|\Delta \vec{v}| = 2v \sin(	heta/2) = 2 	imes 10 	imes \sin(90^\circ/2) = 20 	imes \sin(45^\circ) = 20 	imes (1/\sqrt{2}) = 10\sqrt{2} \approx 14.14\, ms^{-1}$ in the south-west direction.

$A$ scooter going due east at $10\, ms^{-1}$ turns right through an angle of $90^\circ$. If the speed of the scooter remains unchanged in taking the turn,the change in the velocity of the scooter is

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