Question

An athletes runs three rounds on a track of circumference of 200 m. She takes 25 s to run the first round, 40 s for the second round and 35 s for the third round. Then write down the answers to the questions given.
What is the average speed of an athlete? What is the distance travelled by the athlete in the first 15 seconds?​

Answer 1

Answer:

Distance and displacement are two terms that appear to be identical but have quite different meanings and definitions. The term “distance” relates to how much ground an object has covered during its motion, whereas “displacement” refers to how far out of place is an object.

Given parameters

The diameter of the circular track = 200 m

Radius (r) = 200/2

r = 100 m

Time is taken by an athlete to complete one round = 40 seconds.

Displacement = ??

Speed = Distance/Time

Total time athlete moves – 2 min 20 seconds

2 min 20 seconds = 2 × 60 + 20

2 min 20 seconds = 140 seconds

Answer 2

Given: An athletes runs three rounds on a track of circumference of 200 m and takes 25 s to run the first round, 40 s for the second round and 35 s for the third round.

To find:

(i) The average speed of an athlete.

(ii) The distance travelled by the athlete in the first 15 seconds.

Solution:

(i)

• Average speed of the athlete can be calculated using,

        $(v)_{avg} = \frac{d}{t}$

• Here, $(v)_{avg}$ is the average velocity of the athlete, d is the total distance covered by the athlete and t is the total time taken by the athelete.
• So, the total distance would be thrice of 200m, since the athlete runs three rounds.
• Hence, the average velocity of the athlete is,

       $(v)_{avg} = \frac{200m * 3}{25s + 40s + 35s}$

       $(v)_{avg} = \frac{600m}{100s}$

                  $= 6 ms^{-1}$

(ii)

• To find the distance travelled in the first 15 seconds, we need to first find the speed of the athlete during her first round.
• So, the speed of the athlete is,

        $v = \frac{200m}{25s}$

           $= 8ms^{-1}$

• Distance travelled in the first 15 seconds is given by,

        $v = \frac{d}{t}$

        $8 ms^{-1} = \frac{d}{15s}$

       $d = 120m$

Therefore, the average speed of the athlete is 6ms⁻¹ and the distance travelled by the athlete in the first 15 seconds is 120m.