Calculating Snowfall Start Time with Constant Snowplow Rate | Homework Problem

[playboy]

Almost. That is only correct if it started snowing at time $t=0$. But you don't know when it started snowing. The trick is to invent some variable, call it $t_0$ to represent the time at which it started snowing.

I see what your saying.

Would this work:

Let $t_0$ = 12:00 - b
and we must solve for b?

 

[D H]

You still haven't developed an equation for the snow height, $h(t)$. You wrote $h(t) = \dot h t$, where $\dot h$ is the constant snowfall rate. This is not correct, because it yields the snow height at the onset of the storm as $h(t_0) = \dot h t_0$. The height at the onset is presumably zero.

 

[playboy]

You still haven't developed an equation for the snow height, $h(t)$. You wrote $h(t) = \dot h t$, where $\dot h$ is the constant snowfall rate. This is not correct, because it yields the snow height at the onset of the storm as $h(t_0) = \dot h t_0$. The height at the onset is presumably zero.

let $t_0$ be the time it started snowing (and this is what we wan to ultimately solve)

let $\dot h$ be the rate of snow fall.

then $h(t) = \dot h (t-t_0)$ where t>$t_0$.

 

[D H]

Excellent. Now apply this result to the velocity equation, and then integrate velocity (definite integral) wrt time to get the distance traveled over some time interval.

 

[playboy]

So now we have:

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


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

$$v(t) = \frac{\dot V_{\text{plow}}}{w\;h(t)}$$

$h(t) = \dot h (t-t_0)$

Which gives..

$$v(t) = \frac{\dot V_{\text{plow}}}{w\;\dot h (t-t_0))}$$

and ultimately,

$$\int_{12:00}^{13:00} \frac{\dot V_{\text{plow}}}{w\;\dot h (t-t_0))} dt = 6\text{\, km}$$


$$\int_{13:00}^{14:00} \frac{\dot V_{\text{plow}}}{w\;\dot h (t-t_0))} dt = 3\text{\, km}$$

 

[playboy]

And this can be broken down even further

Let d = distance

$$\int_{12:00}^{13:00} \frac{\d}}}{(t-t_0))} dt = 6\text{\, km}$$

$$\int_{13:00}^{14:00} \frac{\d}}}{(t-t_0))} dt = 6\text{\, km}$$

 

[D H]

So evaluate the integrals. They should be easy, since $\dot V_{\text{plow}}$, $w$, $\dot h$, and $t_0$ are all constants.

 

[playboy]

^ Stupid latex... it's supposed to be
^ int(12,13) Distance\(t-to) dt = 6
^ int(13,14 Distance\(t-to) dt = 3

 

[playboy]

So evaluate the integrals. They should be easy, since $\dot V_{\text{plow}}$, $w$, $\dot h$, and $t_0$ are all constants.

Yea...that's what I was trying to write out

 

[playboy]

Let all those constants = D

int(12,13) D\(t-to) dt = 6
=D*(ln(12-to) - ln(13-to)) = 6
=(ln(12-t0) - ln(12-to)) = 6\D
= and doing a whole bunch of math...

to = [e^(6\D)*13 - 12] \ [e^(6/D) - 1]

Similarily,

int(13,14) D\(t-to) dt = 3
= D*(ln(13-to) - ln(14-to)) = 3

Solving for to I get,

to = [e^(3\D)*14 - 12] \ [e^(3/D) - 1]

 

[playboy]

This gets VERY messy...

Would it be the same as letting the limits of the intergral be (0,1) and (1,2) instead of (12, 13) and (13, 14) ?

 

[D H]

OK.

You have a sign error here,
int(12,13) D\(t-to) dt = D*(ln(12-to) - ln(13-to))
and similarly here,
int(13,14) D\(t-to) dt = D*(ln(13-to) - ln(14-to))

You have two equations in two unknowns, D and t0. You don't care about D. How can you eliminate that nasty "D" with one simple algebraic manipulation?

 

[playboy]

Subtract the two equations!

 

[playboy]

Mistake, Divide the two equations!

 

[D H]

Excellent. What is the result? Can you get rid of the logarithms?

 

[playboy]

[D*(ln(12-to) - ln(13-to))] \ [D*(ln(13-to) - ln(14-to))] 6\3

= [(ln(12-to) - ln(13-to))] \ [(ln(13-to) - ln(14-to))] = 2

--> ln(12-to) - ln(13-to) = 2ln(13-to) - 2ln(14-to)

-->ln(12-to) -3ln(13-to) + 2ln(14-to) = 0

= ln[ (12-to)*(14-to)^2 / (13-to)^3 ] = 0

take e of both sides...

[ (12-to)*(14-to)^2 / (13-to)^3 ] = 1

(12-to)*(14-to)^2 = (13-to)^3

ALMOST THEIR!

 

[D H]

You have it right and you are indeed almost there. You should easily be able to finish this off. I have to go again. Good luck.

 

[playboy]

Thank you so much for your help D_H..I really reall really appreciate it...

 

[playboy]

I got 3.43

I was hopping it would come out to a really really nice number..

therefore, it strated snowing at 12-3.43 = 9:43am

 

[D H]

That is not what I got.

The final task is to solve for zero,

$$(12-t_0)(14-t_0)^2 - (13-t_0)^3 = 0$$

Expanding,
$$155 - 25\;t_0 + t_0^2 = 0$$

Which has solutions
$$t_0 = \frac {25 \pm \sqrt 5}2$$

Choosing a positive radical means it started snow after noon; i.e., not a true solution. Thus
$$t_0 = \frac {25 - \sqrt 5}2 = 11 + \frac {3 - \sqrt 5}2 = 11.38 \approx 11\text{:}23$$

 

[playboy]

I made a silly mistake and had

155 -25to +to^2 + 2 to^3 = 0 when it should have been 155 -25to +to^2 + 2

I made a mistake on my arithmetic

Why do you have 11:38 ~ 11:23? why 11:23?

D H, you really helped me thorugh this problem and I really really appreciate it...

Thank you so much for your time and patients :)

I have to admit, this was a very tough question for me...

You must be smart and have really good problem solving skills.

 

[D H]

Thanks for the complement. BTW, 11.38 is the answer in hours. Our normal time-measuring system is not metric. We use days, hours, minutes, seconds,... 11.38 hours is about 11 hours and 23 minutes, or 11:23.

 

[playboy]

Thanks for the complement. BTW, 11.38 is the answer in hours. Our normal time-measuring system is not metric. We use days, hours, minutes, seconds,... 11.38 hours is about 11 hours and 23 minutes, or 11:23.

Oh I see, I should have known that,

Thank you once again for your help,

Problems like these are very very complicated (for me...and as you can see, it took me 2 days to figure it out with your help)

I really appreciate your help