Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at 7.0 m/s . Once free of this area, it speeds up to 13 m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16s to reach a higher cruising speed. Required:
What is the final speed?

 v = 25 m / s

Explanation:

For this exercise we use the relations and kinematics

first part, the train accelerates from v₀ = 7.0 m/s to 13 m/s in a time t = 8.0 s

        v = v₀ + a t

        a =

        a =

        a = 0.75 m / s²

second part. Accelerate again for t = 16 s

         v = v₁ + a t

for this interval the initial velocity is v₁ = 13 m / s

         v = 13 + 0.75  16

         v = 25 m / s

cant without the picture! ?

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)