Question

a car travels along a straight line at a constant speedof 60.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. the average velocity of the entire trip is 30.0 mi/h (a) what is the constant speed with which the car moved during the second distance d?

Answer 1

Time for entire trip = 2d/30 hours
Time for first part of trip = d/60.
Time for second part of trip, is given by:
$\frac{2d}{30}-\frac{d}{60}=\frac{d}{20}$
Therefore the speed during the second distance d is 20 mi/h.

Answer 2

it travel d distance at the speed of 60 mph, so t1 = d/60

it travel another distance with v velocity, so t2 = d/v

t = t1 + t2 
  = d/60 + d/v

noww, average  velocity = total distance / total time

Avg.v  = 2d/(d/60 + d/v)
 Avg. v = 120v /  (v+60)

30     =   120 v / (v+60) 

30v + 1800 = 120 v

1800 = 90 v
  

v = 20

hope this helps