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

Answer:

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 = $5\times 53.80$

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' = $\frac{AQ}{t}$

v' = $\frac{53.48}{53.80} = 1.007 m/s$

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

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$=53.8\times 5=269 m$

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$=\frac{Displacement}{time}$

$=\frac{69}{53.8}=1.28 m/s$

Difference between average velocity and average speed is

5-1.28=3.71 m/s