A particle moves with a constant speed of 5 , | vedclass

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(B) The change in velocity is given by the vector difference $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$.Since the speed is constant,let $v = |\vec{v}_1|

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) The change in velocity is given by the vector difference $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$.Since the speed is constant,let $v = |\vec{v}_1| = |\vec{v}_2| = 5 \, ms^{-1}$.For a half revolution,the angle between the initial velocity vector $\vec{v}_1$ and the final velocity vector $\vec{v}_2$ is $180^\circ$.The magnitude of the change in velocity is given by $\Delta v = |\vec{v}_2 - \vec{v}_1| = \sqrt{v^2 + v^2 - 2v^2 \cos(180^\circ)} = \sqrt{2v^2 + 2v^2} = 2v$.Alternatively,using the formula $\Delta v = 2v \sin(	heta/2)$,where $	heta = 180^\circ$:$\Delta v = 2 	imes 5 	imes \sin(180^\circ / 2) = 10 	imes \sin(90^\circ) = 10 	imes 1 = 10 \, ms^{-1}$.

$A$ particle moves with a constant speed of $5 \, ms^{-1}$ as shown in the figure. What is the change in velocity during half a revolution in $ms^{-1}$?

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