Question

A body is moving with constant speed, in a circle of radius $10 m$ .The body completes one revolution in 4 s .At the end of 3 rd second, the displacement of body (in mi) from its starting point is :

• A. $15 \pi$
• B. $10 \sqrt{2}$
• C. 30
• D. $5 \pi$

Hint:

For circular motion, the displacement from the starting point after \( \frac{3}{4} \) of a revolution is \( r\sqrt{2} \), where \( r \) is the radius.

Answer 1

The Correct Option is B

From the given figure above , we get
Speed, v = constant 
Radius, R = 10 m 
T = Time period = 4s 
At the end of 3rd second, particle will be at D (Starts from A)
\(\therefore\) displacement S = \(\sqrt2R\)
\(=\sqrt2\times10\)
\(=10\sqrt2\)
So, the correect answer is (B) : $10 \sqrt{2}$

Answer 2

Given: The circle of the radius (R) = 10m,
Speed is constant,
Total time given = 4s
Displacement is the shortest distance between the initial and the final position.

From the above diagram it is clear that the initial position of the body is at A and the final position is D.

Total time of 4 seconds is evenly distributed for each segment of the orbit:

• It takes 1 second to travel from point A to point B
• Similarly, it takes 1 second from point B to point C
• Another 1 second is required to travel from point C to point D
• Finally, 1 more second from D to A to complete one revolution.

At the end of the 3rd second, the particle will be at D (when Starts from A). 
As from the figure, it is clear AOD is right angled triangle, applying Pythagoras theorem, Displacement = S
\(S=\sqrt{AQ^2+OD^2}\)
\(S=\sqrt{R^2+R^2}\)
\(S=R\sqrt{2}\)
\(S=10\sqrt{2}\)
Therefore, At the end of 3 rd second, the displacement of body (in mi) from its starting point is \(10\sqrt{2}.\)