Question

1. An athlete covers a circular path of radius 70 m in 555
and returns to the starting point. Find the magnitude of
his average speed and average velocity. If he completes
the last 40 m of the path at the same rate in 4 s, find the
value of instantaneous speed and velocity in that inter-
val of time. [8 m.s-1,0; 10 m.s-1, 10 m.s-1​

Answer 1

Answer:

Given that,

Diameter D=200m

Radius r=100m

Time, complete one round t=40s

Time in 2 minute 20 second t=2×60+20

We know that,

Circumference of circle =2πr=2×3.14×100=628m

Now, in 40 s athlete cover a distance of 628 m

In 1 s distance covered by the athlete =40628=15.7

In 140 s distance covered by the athlete =15.7×140=2198m

Now, the number of rounds

  n=6282198

 n=3.5

So, in given time of 2 minute 20 second, athlete will complete 3.5 rounds to the circular track

 

Now, 3.5 rounds means three rounds and half round. This means after the 3.5 rounds athlete will be just opposite to the starting point that means the displacement is equal to the diameter of circular path

Hence, the distance covered is 2198 m and displacement is 200 m

Answer 2

Answer:

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