SOLUTION: A minivan leaves Topeka at 8:00am. Half an hour later, a car leaves Topeka and follows the same route as the minivan. The car's speed is 5mph faster than the minivan's speed. If th

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Question 1150145: A minivan leaves Topeka at 8:00am. Half an hour later, a car leaves Topeka and follows the same route as the minivan. The car's speed is 5mph faster than the minivan's speed. If the car catches up to the minivan at 3:30 pm, how fast is each vehicle traveling?
Found 2 solutions by ikleyn, josmiceli:
Answer by ikleyn(52864)   (Show Source): You can put this solution on YOUR website!
.

Before the catching moment, the minivan was on the way 7.5 hours, from 8:00 am to 3:30 pm.

                            The car     was on the way half an hour less, i.e. 7.5-0.5 = 7 hours.


Let x be the average speed of the minivan, in miles per hpur.

Then the average speed of the car was (x+5) mph, according to the condition.


To the catching moment, they covered the same distance.  Hence,


    7.5*x = 7*(x+5).


From this equation,


    7.5x = 7x + 35

    7.5x - 7x = 35

    0.5x      = 35

       x      =  = 70.


ANSWER.  The minivan average speed was 70 mph.  That of car was 70+5 = 75 mph.

Solved.


Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Let = the speed of the minivan in mi/hr
= the speed of the car in mi/hr
---------------------------------------
From 8:30AM to 3:30 PM is 7 hrs
The minivan's head start is mi
Let = the distance the car travels from
8:30 until it catches the minivan at 3:30 PM
Start a stopwatch when the car leaves at 8:30
-------------------------------------------------
Equation for the minivan:
(1)
Equation for the car:
(2)
-------------------------------
Plug (2) into (1)
(1)
(1)
(1)
(1)
and

-----------------------
The car's speed is 75 mi/hr
The minivan's speed is 70 mi/hr
-------------------------------------
check answer:
(1)
(1)
(1)
(1)
and
(2)
(2)
(2)
OK

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