Word Problem

Question

During the first part of a trip, Sam's Chevrolet Cobalt traveled 120 mi. Sam then drove another 100 mi at a speed that was 10 mph slower. If the total time for Sam's trip was 4 hr, what was his speed on each part of the trip.

Answer 1

 

Rare Time = Distamce

 

Let        R = rate of first leg of the trip

Let R - 10 = rate of trip at slower speed

 

PART 1 OF TRIP:

 

    Rate = R

    Distance = 120

 

    Since Rate • Time = Distance

             Time = Distance ÷ Rate

                     = 120/R

 

PART 2 OF TRIP:

 

     Rate = R - 10 

     Distance = 100

 

     Since Rate • Time = Distance

              Time = Distance ÷ Rate

                      = 100/(R - 10)

 

Since the total time of the trip was 4 hours, we can construct and solve the following equation:

 

    Time (1st part of trip) + Time (2nd part of trip) = 4

    120/R + 100/(R - 10) = 4

 

LCM = (R)(R - 10)

Multiply all terms by R(R - 10) to clear the fractions

 

    (120/R)(R)(R - 10) + ((100/(R - 10))(R)(R - 10) = 4(R)(R - 10)

    120(R - 10) + 100R = 4R² - 40

    120R - 1200 + 100R = 4R² - 40

    220R - 1200 = 4R² - 40

    0 = 4R² - 220R + 1160

    0 = R² - 55R + 290

    

This will not factor so use the quadratic formula to solve.

 

R = 49.28 miles per hour

 

So he traveled at a rate pf 49.09 mph on the first part of the trep.

He traveled 10 miles per hour slower on the second part ... 39.09 mph