Question

A cyclist is moving with a constant acceleration of $1.2m{s}^{-2}$ on a straight track. A racer is moving on a circular path of the radius $150m$ at a constant speed of $15m{s}^{-1}$ . Find the magnitude of the velocity of racer which is measured by the cyclist has reached a speed of $20m{s}^{-1}$ for the position represented in the figure.

• A. $18.03m{s}^{-1}$
• B. $25m{s}^{-1}$
• C. $20m{s}^{-1}$
• D. $15m{s}^{-1}$

Answer 1

Answer: (A)

Let ${\overrightarrow{v}}_c$ = actual velocity of cyclist along $x-axis$
${\overrightarrow{v}}_r$ = the velocity of racer along tangent of circular path (shown in figure)
$\therefore$ ${\overrightarrow{v}}_{rc}={\overrightarrow{v}}_r-{\overrightarrow{v}}_c$


Hence ${\overrightarrow{v}}_c=20\hat{i}$
${\overrightarrow{v}}_{rc}=V_{r}cos30^o\hat{j}+V_{r}sin30^o\hat{i}$
Or ${\overrightarrow{v}}_r=15cos30\hat{j}+15sin30\hat{i}$
$=\frac{15\sqrt{2}}{2}\hat{j}+\frac{15}{2}\hat{i}$
$\therefore$ ${\overrightarrow{v}}_{rc}={\overrightarrow{v}}_r-{\overrightarrow{v}}_c$
$=\frac{15\sqrt{3}}{2}\hat{j}+\frac{15\sqrt{3}}{2}\hat{j}$
$=-12.5\hat{i}+\frac{15\sqrt{3}}{2}\hat{j}$
$\therefore \left|{\vec{v}}_{rc}\right|=\sqrt{(-12.5{)}^2+{\left(\frac{15\sqrt{3}}{2}\right)}^2}$
$=18.03m{s}^{-1}$