Question

A person starts to run around a square track with sides of 50 m starting in the bottom right corner and running counter-clockwise. Running at an average speed of 5 m/s, the jogger runs for 53.80 seconds. What is the difference between the magnitude of the person's average velocity and average speed in m/s? (please provide detailed explanation)

Answer 1

The difference in average speed and average velocity in terms of magnitude is 3.993 m/s

Solution:

As per the question:

The side of a square track, l = 50 m

Average speed of the runner,
`v_(avg) = 5 m/s`

Time taken, t = 53.80 s

Now,

The distance covered by the runner in this time:

s =
`v_(avg)t`

s =
`5* 53.80`

s = 269 m

After covering a distance of 269 m, the person is at point A:

`AQ^(2) = AR^(2) + QR^(2)`

where

AR = 19 m

QR = 50 m

Refer to fig 1.

As the runner starts from the bottom right, i.e., at Q and traveled 269 m.

After completion of 250 m , he will be at point R after one complete round and thus travels 19 m more to point A to cover 269 m.

Thus

`AQ = \sqrt{19^(2) + 50^(2)} = 53.48 m`

where

AQ is the displacement

Hence,

Average velocity, v' =
`(AQ)/(t)`

v' =
`(53.48)/(53.80) = 1.007 m/s`

The difference in average speed and average velocity is:

`v_(avg) - v' = 5 - 1.007 = 3.993 m/s`

Answer 2

Answer:3.71 m/s

Step-by-step explanation:

Given

square track with sides 50 m

average speed is 5 m/s

Total running time=53.8 s

Total distance traveled in this time
`=53.8* 5=269 m`

i.e. Person has completed square track one time and another 69 m in second round

So displacement is 269-200=69 m

average velocity
`=(Displacement)/(time)`

`=(69)/(53.8)=1.28 m/s`

Difference between average velocity and average speed is

5-1.28=3.71 m/s