# Snow Plow Problem

1. Feb 5, 2017

### Matthew R

• Thread moved from the technical PF forums so no Homework Help Template is shown
Hi guys.

I am currently stuck on the classic snow plow problem. I have the following differential equation and initial conditions:
@ 7am plow starts off to clear snow at a constant rate
By 8am, plow has gone 4mi
By 9am, plow has gone an additional 3mi
Let t=0 when it started to snow, when did it start to snow. They gave the answer as 4:27am

k(dx/dt)=1/t

It wants me to find the constant k.

Here is my work:

∫k dx= ∫dt/t
kx +C = ln (t) At this point I am assuming that x is the independent variable and t is the dependent variable.

e^(kx) +C = t
Using the first initial condition of when it starts to snow of x(0)=0, I get the C is 0, therefore

e^(kx) = t. Using the second initial condition @8, x(t+1)=4

e^(4k)= t+1.

This is the point I get stuck. I cant seem to solve for k that has a real number.

Any help would be appreciated.

2. Feb 6, 2017

Right.
Wrong.

3. Feb 6, 2017

### Matthew R

Ok, looking back then the c would be e^c which would then be Ae^(kx)=t? But having initial condition of t(0)=1 would make A=0.

4. Feb 6, 2017

### haruspex

At t=0, the snow plow was safely in its garage.

5. Feb 6, 2017

### Matthew R

I understand that but will always result in a zero answer if A is 0.

6. Feb 6, 2017

### haruspex

The equation you used to specify its motion is only valid when.... what?

7. Feb 6, 2017

### Matthew R

At t=0. So is A not a factor? Leaving e^kx=t? Then would I just plug in the next conditions and solve for k?

8. Feb 6, 2017

### haruspex

No. t=0 is when it started snowing. At that time, the plow was going nowhere. The plow set off at 7 am. (We do not know when that was as a value of t.) The equation only applies while the plow is moving.

9. Feb 7, 2017

### pasmith

Actually it doesn't; it wants you to find $T$, which is the time from when it started snowing until 7am. You do not need to know $k$ in order to find this.

However I think you may have copied the question incorrectly, because as far as I can tell you end up with a sixth-order polynomial for $T$ which doesn't have a root anywhere near the alleged answer of $T = 2.55$.

10. Feb 7, 2017

### haruspex

Do you mean a sixth order polynomial for ek?