SOLUTION: A train travels from city A to city B and then to city C. The distance from A to B is 51 miles and the distance from B to C is 75 miles. the average velocity from A to B was 35mph

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: A train travels from city A to city B and then to city C. The distance from A to B is 51 miles and the distance from B to C is 75 miles. the average velocity from A to B was 35mph       Log On

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Question 562018: A train travels from city A to city B and then to city C. The distance from A to B is 51 miles and the distance from B to C is 75 miles. the average velocity from A to B was 35mph and the average velocity from B to C was 50mph. What was the average velocity from A to C in miles per hour?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A train travels from city A to city B and then to city C.
The distance from A to B is 51 miles and the distance from B to C is 75 miles.
the average velocity from A to B was 35mph and the average velocity from B to C was 50mph.
What was the average velocity from A to C in miles per hour?
:
The total distance: 51 + 75 = 126 mi
:
Let a = average for the whole trip
:
Write a time equation; time = dist/speed
:
AB time + BC time = AC time
51%2F35 + 75%2F50 = 126%2Fa
Multiply by 350a to clear denominators, results
10a(51) + 7a(75) = 350(126)
510a + 525a = 44100
1035a = 44100
a = 44100%2F1035
a = 42.6 mph av for AC
:
:
Check;
51%2F35 + 75%2F50 = 126%2F42.6
1.457 + 1.5 = 2.947