
#1
Jan2413, 08:17 PM

P: 14

1. The problem statement, all variables and given/known data
Driving along a crowded freeway, you notice that it takes a time t to go from one mile marker to the next. When you increase your speed by 4.5 mi/h, the time to go one mile decreases by 10 s. What was your original speed? 2. Relevant equations V= D/T 3. The attempt at a solution V1= D/T V2= (D+4.5)/(T10) 4.5mph = 6.6ft/s That's all i could come up with so far. Thanks in advance for the help 



#2
Jan2413, 08:21 PM

P: 634





#3
Jan2413, 08:27 PM

P: 14

4.5mph = 6.6ft/s
V1= 1/T V2+6.6ft/s= (1)/(T10) Velocity is used for speed correct? I also realized it gives you distance.. duhhh.. How do i proceed from here? Am i trying to isolate time to but into the v1 equation? 



#4
Jan2413, 08:31 PM

P: 634

Velocity, Speed, and Time questionNotice how the question says "When you increase your speed by 4.5 mi/h" ... it means V2 = V1 + 4.5. How do the equations look now? 



#5
Jan2413, 08:43 PM

P: 14

which means T2= t110 , I think...
I feel like it should be a simple plug in now but i cant seem to find out how. V1+6.6= 1/(t110) I'm missing something.. either a way to get time or velocity so i can solve for the other.. 



#6
Jan2413, 08:51 PM

P: 634





#7
Jan2413, 08:52 PM

P: 501

You have two equations and two unknowns. Solve one of the equations for one of the unknowns, and plug the result into the other equation.




#8
Jan2413, 09:12 PM

P: 14

Ok, so what i think you're saying is...
V1+4.5= 1/T110 v1= (1/t110)4.5 so.. (1/t1104.5) = 1/t? then solve that for T? 



#9
Jan2413, 09:16 PM

P: 634

[itex]V_{1} = \frac{1}{T_{1}10}  4.5[/itex] and the second line doesn't agree with this. It should read [itex]\frac{1}{T_{1}} = \frac{1}{T_{1}10}  4.5[/itex] 



#10
Jan2413, 09:20 PM

P: 14

but don't you plug what i had into the first equation for v1?
i didn't mean to put the parenthesis where i did. i mean to put so. (1/t110)4.5 = 1/t then solve for t. is this still not correct? nvm... 


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