Guadalupe went on a 192 mile trip to a game. due to road construction she had to drive 16 miles per hour slower. This made the trip take 1 hour longer. How fast did she drive to the soccer game?

Answer 1

The total distance is X = 192 miles, from the problem statement the actual velocity is 16 miles/hr lower than than the velocity that would have been traveled. Lets call this velocity V_act and call the would be velocity as V

then V_act = V -16 mile/hr = V -16 ...eq 1

the actual time driven would be 1 hr more , and lets call that time t_act = t + 1 ...eq.2

where t is what would be time if driven at velocity V. The time t = X / V and note that X is the same distance no matter what velocity is , and t_act = X / V_act. Substitute t and t_act in eq 2 and you get:

( X/ V_act ) = ( X / V ) + 1 , but V_act = V -16 , now substitute this into the equation first term and you get:

( X / V -16 ) = ( X / V) +1 , solve for V , when X = 192

( X / V -16 ) - ( X /V) = 1 , and X V - X ( V-16) / V ( V -16 ) = 1, now cross mutiply and you get

XV - X ( V -16 ) = V ( V - 16) , XV - XV + 16 X = V² - 16 V, and 16 X = V² - 16 V, now substitute X

16( 192) = V² - 16 V , 3072 = V² -16V. Rearrange equation, V² - 16 V - 3072 = 0 this is a quadratic equation

with two roots or two possible solutions for V, solving the quadratic equation:

V = 16 ± √ (-16)² - 4(1)(-3072) / 2 , this solution has two roots

V = (16 + 112) / 2 = 64 miles/hr , and V = (16 - 112 ) / 2 = -48 miles/hr , the 2nd solution is struck out because velocity cannot be negative.

Therefore Guadalupe would be velocity is 64 miles /hr and her actual velocity is 64 -16 = 48 miles/hr

solving for would be time t = X / V = 192 miles / 64 miles /hr = 3 hrs, but the actual driving time is 1 hr more

t_act = t + 1 = 3 + 1 = 4 hrs, checking the actual velocity, V_act = X / t_act = 192 miles / 4 hr = 48 miles /hr and the solution checks