Question

A student runs along a circular track of diameter \(80\) m. The distance travelled and the magnitude of displacement of the student when he covers \(\left(\frac{3}{4}\right)^{th}\) of the circle is (in m)

• A. \(60\pi ,40\sqrt{2}\)
• B. \(40\pi ,60\sqrt{2}\)
• C. \(120\pi ,80\sqrt{2}\)
• D. \(80\pi ,120\sqrt{2}\)

Hint:

Distance is the arc length covered, while displacement is the straight chord joining the start and end points.

Answer 1

The Correct Option is A

Step 1: Set up coordinates
Put the centre at the origin and the start at $(R,0)$.
After $270^{\circ}$ the student is at $(0,-R)$.

Step 2: Displacement
$\sqrt{R^2+R^2}=\sqrt{2}\times40=40\sqrt{2}$ m.

Step 3: Distance
Arc length $=R\theta=40\times\dfrac{3\pi}{2}=60\pi$ m, which gives option (A).

Final Answer:
Distance is 60 pi m and displacement is 40 root 2 m. \[ \boxed{\text{(A) }60\pi,\ 40\sqrt{2}} \]