A truck traveled the first 110 miles of a trip at one speed and the last 150 miles at an average speed of 5 miles per hour less. The entire trip took 5 hours. W

Question

A truck traveled the first 110 miles of a trip at one speed and the last 150 miles at an average speed of 5 miles per hour less. The entire trip took 5 hours. What was the average speed for the first part of the trip?

Answer 1

distance = rate x time

 

110 mi = R * (5-T) ---> 110/(5-t) = R

 

150 mi = (R-5)* T = RT - 5T

 

 

Substitutes 1st equation into the second:

150 =  [110/(5-t) - 5/1 ]* T

 

150 = [ 110/(5-T) - 5(5-T)/(5-T)]* T

 

150 =  [  110 - 5(5-t) ]* T/(5-T)

 

150 = [ 110 - 25 + 5T]*T / (5-T)

 

150 = [ 85 + 5T]*T/(5-T)

 

150(5-T) = [ 85 + 5T]*T

 

750 - 150T = 85T + 5T^2

 

0 = 5T^2 + 85T + 150T - 750

 

0 = 5T^2 + 235T - 750

 

0 = T^2 + 47T - 150

 

0 = ( T    +  50   )( T    -   3  )

 

T+50=0  results in negative measures

 

T-3=0 --> T=3

 

Per the first equation

 

110 = R(5-t) = r(5-3) = 2r, r = 110/2 = 55 mph

 

Then 150 = 3r ---> r = 50

 

Truck travels for 2 hours at 55 mph , for a distance of 2x55 = 110 miles.

Then the truck travels for 3 hours at 50 mph for a distance of 150 miles.