In Uniform Circular Motion (U.C.M.),when the | vedclass

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(B) In Uniform Circular Motion,the velocity vector $v$ is always tangent to the circular path. Let the velocity at time $t$ be $v_1$ and at time $t +

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(B) In Uniform Circular Motion,the velocity vector $v$ is always tangent to the circular path. Let the velocity at time $t$ be $v_1$ and at time $t + \delta t$ be $v_2$. The change in velocity is $\delta v = v_2 - v_1$. Since the speed is constant in $U$.$C$.$M$.,$|v_1| = |v_2| = v$. The vector $\delta v$ forms the base of an isosceles triangle with sides $v_1$ and $v_2$ having an angle $	heta$ between them. The angle $\phi$ between the change in velocity $\delta v$ and the initial velocity $v_1$ is given by $\phi = \frac{180^{\circ} - 	heta}{2} = 90^{\circ} - \frac{	heta}{2}$. As the time interval $\delta t ightarrow 0$,the angle $	heta$ between the velocity vectors also approaches $0^{\circ}$. Substituting $	heta \approx 0^{\circ}$ into the expression,we get $\phi = 90^{\circ} - 0^{\circ} = 90^{\circ}$. Thus,the change in velocity is perpendicular to the instantaneous velocity vector.

In Uniform Circular Motion ($U$.$C$.$M$.),when the time interval $\delta t \rightarrow 0$,the angle between the change in velocity $(\delta v)$ and the linear velocity $(v)$ will be (in $^{\circ}$)

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