# Velocity, Speed, and Time question

1. Jan 24, 2013

### 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

2. Jan 24, 2013

### Sourabh N

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

3. Jan 24, 2013

### 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?

4. Jan 24, 2013

### Sourabh N

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?

5. Jan 24, 2013

### OUstudent

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..

6. Jan 24, 2013

### Sourabh N

Yes, that is correct.
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?

7. Jan 24, 2013

### tms

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. Jan 24, 2013

### 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?

9. Jan 24, 2013

### Sourabh N

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
$\frac{1}{T_{1}} = \frac{1}{T_{1}-10} - 4.5$

10. Jan 24, 2013

### OUstudent

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...

Last edited: Jan 24, 2013
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