Question

A car is moving round a circular track with a constant speed $v$ of $20m{s}^{-1}$ (as shown in figure). At different times, the car is at $A,B,C$ and $D$, respectively. Find the velocity change

(a) from $A$ to $C$, and

(b) from $A$ to $B$

Answer 1

(a). Change in velocity, as the particle moves from $A$ to $C$, is the difference of final velocity vector and initial velocity vector.

$\Delta {\overrightarrow{v}}_{AC}={\overrightarrow{v}}_C$ - ${\overrightarrow{v}}_A$ = ${\overrightarrow{v}}_C+(-{\overrightarrow{v}}_A)$

(If we take left direction as positive, the right direction will be negative.)

$\Delta {\overrightarrow{v}}_{AC}=20+20=40m{s}^{-1}$

As $\Delta {\overrightarrow{v}}_{AC}$ is positive, hence, change in the velocity should be in the direction of left, i.e., in the direction of ${\overrightarrow{v}}_C$

$\Delta {\overrightarrow{v}}_{AC}=40m{s}^{-1}$ in the direction of ${\overrightarrow{v}}_C$

(b). Change in velocity, as the particle moves from $A$ to $B$,

$\Delta {\overrightarrow{v}}_{AB}={\overline{v}}_B-{\overrightarrow{v}}_A={\overrightarrow{v}}_B+(-{\overrightarrow{v}}_A)$

Since velocities at A and B are perpendicular to each other so

$\Delta v_{AB}=\sqrt{{20}^2+{20}^2}=\sqrt{800}m{s}^{-1}$

$\tan \theta =\frac{20}{20}=1$; $\theta ={45}^{\circ }$

So this change in velocity will be parallel to line joining BC