SOLUTION: A car travels along a straight line at a constant speed of 62.5 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average veloc

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Question 712328: A car travels along a straight line at a constant speed of 62.5 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 32.5 mi/h.
I don't get how to set up this problem at all..

Found 2 solutions by ankor@dixie-net.com, stanbon:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A car travels along a straight line at a constant speed of 62.5 mi/h for a distance d and then another distance d in the same direction at another constant speed.
The average velocity for the entire trip is 32.5 mi/h.
:
Let s = "another constant speed"
:
Write a time equation; time = dist/speed
:
1st dist time + 2nd dist time = total time
d%2F62.5 + d%2Fs = %282d%29%2F32.5
Multiply equation by 2031.25s (62.5*32.5),results:
32.5sd + 2031.25d = 62.5(2sd)
32.5sd + 2031.25d = 125sd
divide thru by d
32.5s + 20321.25 = 125s
2031.25 = 125s - 32.5s
2031.25 = 92.5s
s = 2031.25%2F92.5
s = 21.96 ~ 22 mph 0n the 2nd d
:
:
Check this using d = 100, see if it is true
100%2F62.5 + 100%2F22 = %28200%29%2F32.5

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A car travels along a straight line at a constant speed of 62.5 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 32.5 mi/h.
============
1st leg DATA:
distance = d miles ; rate = 62.5 mph ; time = d/r = d/62.5 hours
-----
2nd leg DATA:
distance = d miles ; rate = r mph ; time = d/r hrs
--------
Equation:
average speed = (total distance)/(total time) = (2d miles)/[(d/62.5)+(d/r)]
----
(2d)/[rd+62.5d)/(62.5r)] = 32.5 mph
---
(125rd = 32.5[rd+62.5d]
125rd = 32.5rd + 2031.25d
92.5rd - 2031.25d = 0
etc.
============
Cheers,
Stan H.