Distance/Rate/Time Word Problems
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Distance/Rate/Time Word Problems

[From: ] [author: ] [Date: 11-08-09] [Hit: ]
It took 0.6 hours longer to go there than it did to come back. The average speed on the trip there was 46 km/h. The average speed on the way back was 52 km/h. How many hours did the trip there take?-Hello,......
Jimmy drove to his cabin on the lake and back. It took 0.6 hours longer to go there than it did to come back. The average speed on the trip there was 46 km/h. The average speed on the way back was 52 km/h. How many hours did the trip there take?

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Hello,

You asked questions 2 days ago. You have received right answers. Why not give someone the 10 points ? It would be fair. It costs nothing. No point lost. Ok ? Thank you.

For this question.

V=D/T --- D=VT --- T=D/V

V1 is the speed on the way there
V2 is the speed on the way back. V2 = 52

D1 = D2 = D

T1 = T2 + 0.6

V1 = D/T1 = 46 ==> D = 46T1

Find T1 ?

T1 = T2 + 0.6
T2 = D/V2 = D/52
T1 = D/52 + 0.6
T1 = 46T1/52 + 0.6
52T1 = 46T1 + 31.2
6T1 = 31.2
T1 = 5.2 hours ===> Answer

Hope it helped

Bye !

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r*t=d
46(t+.6)=d
52(t)=d
46(t+.6)=52t
46t+27.6=52t
27.6=6t
t=4.6

Total hours=trip there+trip home
T=4.6+4.6+.6
T=9.8 hours
1
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