# Calc Problem

1. Feb 10, 2007

### playboy

1. The problem statement, all variables and given/known data

Snow began to fall during the morning of February 2 and continued steadily into the afternoon.
At noon a snowplow began removing snow from a road at a constant rate. The plow traveled 6 km from noon to 1 P.M. but only 3 km from 1 P.M. to 2 P.M. When did the snow begin to fall?

2. Relevant equations

t = time

x(t) = distance

dx\dt = speed of the plow

b - the number of hours before noon that it began to snow

3. The attempt at a solution

I know that the speed of the plow is (6 + 3) /2 hours = 4.5km\hour

I am trying to figure out an expression for the height of the snow at time t and then use the given ingormation that the rate of removal R is constant, and thus, find the time it began to snow

Can somebody help me out please?

2. Feb 10, 2007

### D H

Staff Emeritus
The average speed of the plow isn't going to help much here, as the plow's speed is changing (and doing so constantly). This is a calculus problem.

How does the plow's instantaneous speed relate to the constant rate at which the plow removes snow from the road?

How would you interpret the phrase "Snow began to fall during the morning of February 2 and continued steadily into the afternoon" mathematically?

3. Feb 10, 2007

### playboy

I am more lost than ever.

I dont think the instantaneous speed has an anything to do with the problem correct?
The plow could have been going really fast at one point and perphaps at rest at another point?

This simply means that the snow fell at a constant rate from morning into afternoon,
we can call this rate D.

4. Feb 10, 2007

### D H

Staff Emeritus
The instantaneous speed has everything to do with the problem. You are given two distances, both of which are definite integrals of the instantaneous speed.

Formulate these integrals and you will be close to having solved the problem. You will need to have an expression for the instantaneous speed to formulate these integrals.

Last edited: Feb 10, 2007
5. Feb 10, 2007

### playboy

Distance 1, 6km, is on the interval [0,1]

Distance 2, 3 km, is on the interval [1,2]

int(0,1) 6 dt + int(1,2) 3 dt = 9

How does that sound?

6. Feb 10, 2007

### D H

Staff Emeritus
Not very good. You are integrating distance. You need to integrate velocity,

$$\int_{12:00}^{13:00} v(t) dt = 6\text{\, km}$$

$$\int_{13:00}^{14:00} v(t) dt = 3\text{\, km}$$

Now you just have to come up with an expression for $v(t)$. Hint: You know that the removal rate is a constant. How does this relate to velocity and the amount of snow on the ground?

7. Feb 10, 2007

### playboy

The removal rate is Constant.

Therefore, the velocity of the plow is the same as the velocity of snowfall?

8. Feb 10, 2007

### D H

Staff Emeritus
The removal rate is constant, but the velocity is not. Snow is falling all afternoon long, and it had a head start on the snowplow.

9. Feb 10, 2007

### D H

Staff Emeritus
How does the height of the accumulated snow on the road affect the velocity at which the plow can move?

10. Feb 10, 2007

### D H

Staff Emeritus
Another hint: What units would you use to reflect the constant rate at which snow is falling on the road and the constant rate at which the plow is removing snow from the road?

11. Feb 10, 2007

### playboy

I'm confused between "removal rate" and "velocity" ... I thought they were equivalent.

The higher the hight, the smaller the velocity.

Units for snow falling on road --> $$m^2/hour$$ or perhaps $$km^2/hour$$
Units for snow removal --> $$m^3/hour$$ or perhaps $$km^3/hour$$

12. Feb 10, 2007

### D H

Staff Emeritus
Good. Now you're getting somewhere.

Try again. Think of how the weatherman expresses snowfall rates.

Good. Obviously, the snow removal rate cannot equal the snowfall rate since they have different units.

Now factor in the time-varying height of the snow and the fixed width of the plow to determine the velocity of the plow.

Kudos on attempting to learn LaTeX. Nothing beats it (and certainly not ASCII math) for making math look nice.

Last edited: Feb 10, 2007
13. Feb 10, 2007

### playboy

Units for snow falling on road ---> $$m/hour$$

Yeah, Latex is awesome, Something I really have to master

14. Feb 10, 2007

### D H

Staff Emeritus
Excellent. Now for the velocity of the plow ...

15. Feb 10, 2007

### D H

Staff Emeritus
Have to take a break. Home duties are calling. You're almost there. You should get to a point where you have two expressions involving a lot of unknowns, including the two rates, the width of the plow, and time at which it started snow. A simple algebraic manipulation will eliminate all but one (the time). Solve for the time, and voila, you have the answer.

16. Feb 10, 2007

### playboy

:)

Thank you so much for your time! I really appreciate it,

I am still working on this problem..

17. Feb 10, 2007

### playboy

So here is what I have so far:

$$\int_{12:00}^{13:00} v(t) dt = 6\text{\, km}$$

$$\int_{13:00}^{14:00} v(t) dt = 3\text{\, km}$$

RATE OF SNOW REMOVAL is $$km^3/hour$$/hour

RATE OF SNOW FALL is $$km/hour$$/hour

Time of snowfall is $$b$$ hours BEFORE noon.

INSTANTANEOUS veolocity of the plow, $$v(t)$$ ... i still cannot find.

Is the "time-varying height of the snow" the "rate of snowfall" $$km\hour$$ ? The snow did have a head start to fall, so should something be added?

I suppose the width of the plow could just be a constant, which I will call $$w$$

Last edited by a moderator: Feb 10, 2007
18. Feb 10, 2007

### D H

Staff Emeritus
Good. In a short span of time $\Delta t$, what is the volume of snow $\Delta v_{\text{snow}}$ removed by the plow? (Assume a constant velocity $v(t)$ over this short span of time.) How does this relate to the known rate at which the plow removes snow, and what happens in the limit $\Delta t \to 0$?

19. Feb 10, 2007

### playboy

Im very unsure of this...I thought it...and all I can come up with is that the plow can remove only $$x*km^3$$ of snow...

I don't really know how to fit time into it unless i turn it into a rate.

20. Feb 10, 2007

### D H

Staff Emeritus
Between time $t$ and $t+\Delta t$, the plow will have moved about $v(t)\Delta t$ meters down the road. Now suppose the snow height at this time is $h(t)$. Since the plow has a width $w$, it will have removed a volume of snow $\Delta V_{\text{snow}} = w\; h(t)v(t)\Delta t$.

How can you relate this to the plow's snow removal rate, call it $\dot V_{\text{plow}}$?