Average rate between a car and a cyclist

Question

Suppose that a cyclist began a 297 mi ride across a state at the western edge of the? state, at the same time that a car traveling toward it leaves the eastern end of the state. If the bicycle and car met after 4.5 hr and the car traveled 38.6 mph faster than the? bicycle, find the average rate of each.

Answer 1

 

Let x = rate of bicycle

Let x + 38.6 = rate of car

 

The time for both is 4.5 hours

 

RATE • TIME = DISTANCE

 

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For the CAR:

 

   rate • time = distance

   (x + 38.6)•(4.5) = distance

   distance = 4.5x + 173.7

 

For the BICYCLE:

 

   rate • time = distance

   x • 4.5 = distance

   distance = 4.5x

 

Since the total distance traveled is 297, we can construct a 1-variable equation using the sum of the distances for each.

 

   4.5x + 4.5x + 173.7 = 297

   9x = 297 - 173.7

   9x = 123.3

     x = 13.7

 

The average speed of the bicycle is 13.7 miles per hour.

 

    13.7 + 38.6 = 52.3

 

The average speed of the car is 52.3 miles per hour. 

 

    

Answer 2

Work with distance = speed*time

 

d = 297 mi

 

v = speed of cyclist 

u = speed of car = v + 38.6

 

t = time they meet = 4.5 hr

 

They meet at location x = vt = d-ut:

 

vt = d-ut

 

(v)(4.5) = 297-u(4.5)

v(4.5) = 297 - (v+38.6)(4.5)

 

Get v from bottom equation:

 

4.5v = 297 - 4.5v - 173.7

 

9v = 123.3

 

v = 123.3/9 = 13.7 mph

 

u = v+38.6 = 13.7+38.6 = 52.3 mph

 

Test answer:

 

vt = 13.7*4.5 = 61.65

d-ut = 297 - 52.3*4.5 = 61.65

 

So vt = d-ut and we are good.

 

Final answer:

 

Speed of cyclist:  13.7 mph

Speed of car:  52.3 mph