# Velocity, Speed, and Time

1. Feb 10, 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

V1T1 = 1 mi
v2t2= i mile
v2= v1 + change of V
t2= t1 - change of t
I then used equations 3 and 4 to eliminate v2 and t2 in equation 2.
(v1 + change of V)(t1 - change of t) = 1 mi

I know that i need to get rid of t1 now, and i should probably use the first equation (v1t1 = 1 mi) to do it. I just dont know how to integrate the two of them.
Thank you in advance

2. Feb 10, 2013

### tms

You have two equations and two unknowns: $v_1$ and $t_1$. Solve them in the usual way: solve one for one of the unknowns and substitute the result in the other equation.

3. Feb 10, 2013

### OUstudent

That's what i thought i was doing. by putting in v1t1=1mile into (v1 + change in v)(t1-change in t). but would that be (v1 + change in v)(v1 - change in t)? or (v1 + change in v)((1/v1) - change in t)?

4. Feb 10, 2013

### Staff: Mentor

The second one (if you add units), as V1T1=1mile is equivalent to T1=1mile/V1.

5. Feb 10, 2013

### OUstudent

Ah ok. so then i just simplify that equation to find v1.
which would be.. (v12)(Δt) + (ΔvΔt) - (Δv)(1 mi) = 0

so then i do quadratic?

6. Feb 10, 2013

### Staff: Mentor

That is a good idea, indeed. But the units of the last equation are wrong, there is some error in the calculation.

7. Feb 10, 2013

### OUstudent

(Δt)v12 + (Δv)(Δt)v1 - (Δv)(1mi) = 0

fixed???

8. Feb 10, 2013

### Staff: Mentor

Now the units match. I didn't check the equations, but you can check the result yourself afterwards with known v1 and t1.

9. Feb 10, 2013

### OUstudent

i got the right answer.
thank you