To solve this problem we will apply the kinematic equations of linear motion, the net changes in distance and time and the definition of speed, which is understood as the path taken by a piece in a certain amount of time. So we have to
PART A) The expression for the average speed x in terms tal distance [tex]d_T[/tex] and the time [tex]t_T[/tex] is
[tex]x = \frac{d_T}{t_T}[/tex]
The total distance is given by
[tex]d_T = d+d'[/tex]
Here d is the distance covered in 13s and d' is the distance covered in 5s
Substitute 126m for d and 63m for d' in above equation as follows,
[tex]d_T = 126m+63m[/tex]
[tex]d_T = 189m[/tex]
The total time is given by,
[tex]t_T = t+t'[/tex]
Here t is the time taken cover 126m and t' is the time taken to cover half way distance.
[tex]t_T = 13s +5s[/tex]
[tex]t_T = 18s[/tex]
Substitute 189m for [tex]d_T[/tex] and 18 for [tex]t_T[/tex] in the previously equation and we have,
[tex]x = \frac{189}{18}[/tex]
[tex]x = 10.5m/s[/tex]
PART B) The expression for the average velocity v is,
[tex]v = \frac{\Delta x}{\Delta t}[/tex]
[tex]\Delta x = x_2 - x_1[/tex]
[tex]\Delta x = 126m-63m[/tex]
[tex]\Delta x = 63m[/tex]
The change in time is given by
[tex]\Delta t = t+t'[/tex]
Here t is the time taken cover 126m and t' is the time taken to cover half way distance.
Substitute 13s for t and 5 for t' in above equation as follows,
[tex]\Delta t = 13s+5s[/tex]
[tex]\Delta t = 18s[/tex]
Therefore the average velocity would be
[tex]v = \frac{63m}{18s}[/tex]
[tex]v = 3.5m/s[/tex]
