Answer:
4.0 m/s
Explanation:
In the first part of the run, the athlete runs a distance of
[tex]d_1 = 300 m[/tex]
at a speed of
[tex]v_1 = 3.0 m/s[/tex]
So, the time he/she takes is
[tex]t_1 = \frac{d_1}{v_1}=\frac{300}{3.0}=100 s[/tex]
In the second part of the run, the athlete covers an additional distance of
[tex]d_2 = 300 m[/tex]
with a speed
[tex]v_2 = 6.0 m/s[/tex]
So, the time taken in this second part is
[tex]t_2 = \frac{d_2}{v_2}=\frac{300}{6.0}=50 s[/tex]
So, the total distance covered is
d = 300 m + 300 m = 600 m
And the total time taken
t = 100 s + 50 s = 150 s
Therefore, the average speed for the entire trip is
[tex]v=\frac{d}{t}=\frac{600}{150}=4.0 m/s[/tex]