Can anyone please help me with the steps

A) v_average = - 10 km / h,  B)    v = 1.6 m / s, v = 17.6 m / s

Explanation:

A) the average speed is the average speed of a body, if we assume that the direction of going up the hill is positive

    v₁ = 40 km / h

    v₂ = - 60 km / h

the average speed is

         v_average =

         v_average = ( 40 - 60)/2

         v_average = - 10 km / h

B) in this case they indicate the acceleration a = 3.2 m / s² and the velocity vo = 9.6 m / s

i) the speed for 2.5 s above

         v = v₀ + a t

as the time is earlier

          t = - 2.5 s

we substitute

          v = 9.6 - 3.2 2.5

          v = 1.6 m / s

ii) the velocity for a subsequent time of 2.5 s

          t = 2.5 s

we substitute

          v = 9.6 + 3.2 2.5

          v = 17.6 m / s

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s