Question

An athlete completes one round of a circular track of radius R in 80 sec. What will be his displacement at the end of 4 min 60 second? . 2πR b. 3лR c. √2R d. Zero
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Answer 1

Answer:

displacement is 1.414 R metres =square root2

Answer 2

Answer:

The displacement of the athlete at the end of 4 min 60 seconds will be √2R i.e.option(c).

Explanation:

Time is taken to complete one revolution=80 seconds

If the particle completes a round of the circular track then its displacement is zero because it comes back to its original position or it comes back to the point from where it had started its motion.

So, displacement in 80 seconds will be zero.

Now,

1 second=1/80 revolution         (1)

The time that is given in the question is,

t=4 min 60 seconds=240+60=300 seconds

For 240 seconds three rounds of the circular track had been completed. So, its displacement will be zero.

Now to find the number of rounds for the rest 60 seconds we will use equation (1);

$60 seconds=\frac{60}{80}=\frac{3}{4}rounds$                 (2)

Angle turned in one complete round is given as,

$\theta=360\textdegree$

The angle turned in 3/4 round is given as,

$\theta'=\frac{3}{4}\times 360\textdegree=270\textdegree=\frac{3\pi}{2}$              (3)

$Displacement=2Rsin\frac{\theta'}{2}$                (4)

$\pi=180\textdegree$                                         (5)

$sin(90\textdegree+\theta)=cos\theta$                      (6)

By substituting the required values in equation (4) we get;

$Displacement=2Rsin\frac{3 \times \pi}{2\times 2}$

$Displacement=2Rsin135\textdegree=2Rsin(90\degree+45\textdegree)$

$Displacement=2Rcos45\textdegree=2R\frac{1}{\sqrt{2}}=\sqrt{2} R$

Hence, the displacement of the athlete at the end of 4 min 60 seconds will be √2R i.e.option(c).