Answer:
1.22 m/s
1.76 m/s
0
1.46 m/s
Explanation:
*in the first part i have taken 26 m as in the part above it is mentioned distance is 26 m.
[tex]s_1[/tex] = Displacement of the first run = 26 m
[tex]s_2[/tex] = Displacement of the second run = 26 m
[tex]t_1[/tex] = Time taken on the first run = 20.7 s
[tex]t_2[/tex] = Time taken on the second run = 14.8 s
Velocity is given by
[tex]v_1=\dfrac{s_1}{t_1}\\\Rightarrow v_1=\dfrac{26}{20.7}\\\Rightarrow v_1=1.26\ \text{m/s}[/tex]
Average velocity of the first run is 1.22 m/s
[tex]v_2=\dfrac{s_2}{t_2}\\\Rightarrow v_2=\dfrac{26}{14.8}\\\Rightarrow v_1=1.76\ \text{m/s}[/tex]
Average velocity of the first run is 1.76 m/s
Displacement for the round trip is 0 as she started and ended at the same point so her average velocity for the round trip is 0 m/s.
Average speed is given by
[tex]s=\dfrac{\text{Total distance}}{\text{Total time}}\\\Rightarrow s=\dfrac{26+26}{20.7+14.8}\\\Rightarrow s=1.46\ \text{m/s}[/tex]
The average speed for the round trip is 1.46 m/s.