The speed of a particle changes from sqrt{5} | vedclass

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(D) Let the initial velocity be $\vec{v}_i$ and the final velocity be $\vec{v}_f$. The magnitudes are given as $|\vec{v}_i| = \sqrt{5} \ m/s$ and $|\v

Vedclass · Accepted Answer

$90$

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(D) Let the initial velocity be $\vec{v}_i$ and the final velocity be $\vec{v}_f$. The magnitudes are given as $|\vec{v}_i| = \sqrt{5} \ m/s$ and $|\vec{v}_f| = 2\sqrt{5} \ m/s$.The magnitude of the change in velocity is given by $|\Delta \vec{v}| = |\vec{v}_f - \vec{v}_i| = 5 \ m/s$.Using the law of vector subtraction,the magnitude of the change in velocity is given by:$|\Delta \vec{v}|^2 = |\vec{v}_f|^2 + |\vec{v}_i|^2 - 2|\vec{v}_f||\vec{v}_i| \cos 	heta$,where $	heta$ is the angle between $\vec{v}_i$ and $\vec{v}_f$.Substituting the given values:$5^2 = (2\sqrt{5})^2 + (\sqrt{5})^2 - 2(2\sqrt{5})(\sqrt{5}) \cos 	heta$$25 = 20 + 5 - 2(2 	imes 5) \cos 	heta$$25 = 25 - 20 \cos 	heta$$0 = -20 \cos 	heta$$\cos 	heta = 0$$	heta = 90^{\circ}$.Thus,the angle between the initial and final velocities is $90^{\circ}$.

The speed of a particle changes from $\sqrt{5} \ m/s$ to $2\sqrt{5} \ m/s$ in a time $t$. If the magnitude of change in its velocity is $5 \ m/s$,the angle between the initial and final velocities of the particle is (in $^{\circ}$)

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