Solve Snow Plow Problem: Find Constant k

Click For Summary

Homework Help Overview

The discussion revolves around a differential equation related to a snow plow problem, where participants are attempting to find a constant k based on given initial conditions and the motion of the plow over time.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are analyzing the differential equation k(dx/dt)=1/t and discussing the implications of initial conditions on the solution. There are attempts to manipulate the equation and clarify the roles of variables and constants.

Discussion Status

Some participants are questioning the correctness of initial assumptions and the interpretation of constants in the equation. There is a recognition that the problem may not require finding k directly to determine the time from when it started snowing until 7 am.

Contextual Notes

There are discussions about the validity of the equation at specific times and the implications of initial conditions, particularly concerning the motion of the plow and the timing of events.

Matthew R
Messages
7
Reaction score
0
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 can't seem to solve for k that has a real number.

Any help would be appreciated.
 
Physics news on Phys.org
Matthew R said:
kx +C = ln (t)
Right.
Matthew R said:
e^(kx) +C = t
Wrong.
 
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.
 
Matthew R said:
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.
At t=0, the snow plow was safely in its garage.
 
I understand that but will always result in a zero answer if A is 0.
 
Matthew R said:
I understand that but will always result in a zero answer if A is 0.
The equation you used to specify its motion is only valid when... what?
 
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?
 
haruspex said:
The equation you used to specify its motion is only valid when... what?
Matthew R said:
At t=0.
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.
 
Matthew R said:
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.

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
pasmith said:
sixth-order polynomial for T
Do you mean a sixth order polynomial for ek?
pasmith said:
which doesn't have a root anywhere near the alleged answer of T=2.55
The root of interest is about 1.09, giving T about 2.55.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
6
Views
3K
Replies
8
Views
1K
Replies
1
Views
2K