Velocity, Speed, and Time question

by OUstudent
Tags: physic help, speed, time, velocity
 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)/(T-10) 4.5mph = 6.6ft/s That's all i could come up with so far. Thanks in advance for the help
P: 634
 Quote by OUstudent 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)/(T-10) 4.5mph = 6.6ft/s That's all i could come up with so far. Thanks in advance for the help
Look at the line I have boldfaced. The question says the speed increases by 4.5 mph, but you're increasing your distance D by 4.5
 P: 14 4.5mph = 6.6ft/s V1= 1/T V2+6.6ft/s= (1)/(T-10) 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?
P: 634

Velocity, Speed, and Time question

 Quote by OUstudent 4.5mph = 6.6ft/s V1= 1/T V2+6.6ft/s= (1)/(T-10) 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?
Yes, here you can use velocity and speed interchangeably.

Notice 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?
 P: 14 which means T2= t1-10 , 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/(t1-10) I'm missing something.. either a way to get time or velocity so i can solve for the other..
P: 634
 Quote by OUstudent which means T2= t1-10 , I think...
Yes, that is correct.
 I feel like it should be a simple plug in now but i cant seem to find out how. V1+6.6= 1/(t1-10) I'm missing something.. either a way to get time or velocity so i can solve for the other..
I think instead of ft/s, you should stick with miles per hour. Then, you will have two equations in two variables! Do you see it?
 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.
 P: 14 Ok, so what i think you're saying is... V1+4.5= 1/T1-10 v1= (1/t1-10)-4.5 so.. (1/t1-10-4.5) = 1/t? then solve that for T?
P: 634
 Quote by OUstudent Ok, so what i think you're saying is... V1+4.5= 1/T1-10 v1= (1/t1-10)-4.5 so.. (1/t1-10-4.5) = 1/t? then solve that for T?
The boldface part is not right. What you have in the first line is
$V_{1} = \frac{1}{T_{1}-10} - 4.5$
and the second line doesn't agree with this. It should read
$\frac{1}{T_{1}} = \frac{1}{T_{1}-10} - 4.5$
 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/t1-10)-4.5 = 1/t then solve for t. is this still not correct? nvm...

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