Respuesta :
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[tex]=53.8\times 5=269 m[/tex]
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[tex]=\frac{Displacement}{time}[/tex]
[tex]=\frac{69}{53.8}=1.28 m/s[/tex]
Difference between average velocity and average speed is
5-1.28=3.71 m/s
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, [tex]v_{avg} = 5 m/s[/tex]
Time taken, t = 53.80 s
Now,
The distance covered by the runner in this time:
s = [tex]v_{avg}t[/tex]
s = [tex]5\times 53.80[/tex]
s = 269 m
After covering a distance of 269 m, the person is at point A:
[tex]AQ^{2} = AR^{2} + QR^{2}[/tex]
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
[tex]AQ = \sqrt{19^{2} + 50^{2}} = 53.48 m[/tex]
where
AQ is the displacement
Hence,
Average velocity, v' = [tex]\frac{AQ}{t}[/tex]
v' = [tex]\frac{53.48}{53.80} = 1.007 m/s[/tex]
The difference in average speed and average velocity is:
[tex]v_{avg} - v' = 5 - 1.007 = 3.993 m/s[/tex]
