A man is tanning around a circular track that is 200 m in ci

A man is tanning around a circular track that is 200 m in circumference. An observer uses a stopwatch its record the runner\'s time at the end of each lap, obtaining the data in the following table. What was the man\'s average speed (rate) between and 80 s and 181 s? What was the man\'s average speed between 314 s and 466 s? Calculate the man\'s speed for each lap. Is he slowing down, seceding up, or neither? slowing down speeding up nether

Solution

a) avearge speed = distance / time

= distnce travelled b/w 80 s and 181 s / time difference b/w 80s and 181s

= (800 m -400m) / (181s-80s)

= 400m / 101 s

= 3.96 m/s

b)

avearge speed = distance / time

= distnce travelled b/w 314 s and 466 s / time difference b/w 314 s and 466 s

= (1600 m -1200m) / (466 s-314 s)

= 400m / 152 s

= 2.63 m/s

c) speed = distance/time

1st lap = 200 m/ 39 s = 5.13 m/s

2nd lap = 400m - 200m / 80 s - 39 s = 200 m/ 41 s = 4.87 m/s

3rd lap = 600m - 400m / 125 s - 80 s = 200 m/ 45 s = 4.44 m/s

4th lap = 800m - 600m / 181 s - 125 s = 200 m/ 44 s = 4.54 m/s

5th lap = 1000m - 800m / 242 s - 181 s = 200 m/ 61 s = 3.28 m/s

6th lap = 1200m - 1000m / 314 s - 242 s = 200 m/ 72 s = 2.77 m/s

7th lap = 1400m - 1200m / 388 s - 314 s = 200 m/ 74 s = 2.70 m/s

8th lap = 1600m - 1400m / 466 s - 388 s = 200 m/ 78 s = 2.56 m/s

Hence, spped is decreased from 5.13 m/s to 2.56 m/s.

Therefore,

answer = slowing down