KCET · Physics · Motion In One Dimension
An athlete runs along a circular track of diameter 80 m . The distance travelled and the magnitude of displacement of the athlete when he covers \(3 / 4\) 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 \(80 \pi, 120 \sqrt{2}\)
Answer & Solution
Correct Answer
(A) \(60 \pi, 40 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
Given, diameter, \(d=80 \mathrm{~m}\)
\(\therefore \quad r=40 \mathrm{~m}\)

Distance travelled after completion of \(\frac{3}{4}\) revolution
\(=\frac{3}{4} \times 2 \pi r=\frac{3 \pi}{2} \times 40=60 \pi \mathrm{~m}\)
Displacement \(=\sqrt{r^2+r^2}=r \sqrt{2}=40 \sqrt{2} \mathrm{~m}\)
\(\therefore \quad r=40 \mathrm{~m}\)

Distance travelled after completion of \(\frac{3}{4}\) revolution
\(=\frac{3}{4} \times 2 \pi r=\frac{3 \pi}{2} \times 40=60 \pi \mathrm{~m}\)
Displacement \(=\sqrt{r^2+r^2}=r \sqrt{2}=40 \sqrt{2} \mathrm{~m}\)
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