What time did the snow start? Snow Plow I am having trouble proceeding with the problem below. I have solved some stuff to a certain extent, but do not understand what to do from here. The problem statement is: 
One morning snow began to fall at a heavy and constant rate. A snowplow starts out at 8:00am and at 9:00am it has traveled 2 miles. By 10:00am it has traveled 3 miles. Assuming that the plow removes a constant volume of snow per hour, determine the time at which it started to snow. 
Hint: Let t denote the time since the snow started and T be the time when the snowplow started out. Let x be the distance the snowplow has traveled, and h  the height of the snow which is a function of t . Assuming a constant volume of snow per hour is removed implies the speed of the plow times the height of the snow is a constant. Set up and solve differential equations involving dx/dt and dh/dt. 
My work so far: 
$$ \frac {dh}{dt} = C$$
$$ \int {\frac {dh}{dt}} = \int {C}$$
$$ h(t) = Ct+Z $$
$$\frac {dx}{dt} = \frac E{h(t)}$$
$$\frac {dx}{dt} (Ct+Z) = E$$
Separating and integrating this equation, I get :
$$x(t) = (\frac EC) \ln |Ct+Z| + F$$
I believe the the conditions are : x(0) = 0, x(1) = 2, x(2) = 3 
 A: Not quite. Remember that we set $t$ to be the time it started snowing, not when the snow plow started out. So you have $h(0) = 0$, which solves $Z = 0$, so $h(t) = Ct$. 
Putting that in the second equation:
$$\frac{dx}{dt}(Ct) = E$$
$$ \frac{dx}{dt} = \frac{E}{Ct} = \frac{A}{t} $$
which solves
$$ x(t) = A\ln t + K $$
Since $T$ is the time the snow plow started (also what question was asking for; always check what the question asks for), you have $x(T) = 0$, $x(T+1) = 2$ and $x(T+2) = 3$. You now have 3 equations in 3 unknowns - $A$, $K$, $T$
How to solve:
$$ A\ln(T) + K = 0 \tag{1} $$
$$ A\ln(T+1) + K = 2 \tag{2} $$
$$ A\ln(T+2) + K = 3 \tag{3} $$
Subtracting $(1)$ from $(2)$ gives
$$ A\big(\ln(T+1) - \ln(T)\big) = 2 \tag{4} $$
Subtracting $(3)$ from $(1)$ gives
$$ A\big( \ln(T+2) - \ln(T)\big) = 3 \tag{5} $$
Combining $(4)$ and $(5)$:
$$ \frac{A\big(\ln(T+1) - \ln(T)\big)}{2} = \frac{A\big(\ln(T+2) - \ln(T)\big)}{3} $$
$$ 3\big(\ln(T+1) - \ln(T)\big) = 2\big(\ln(T+2) - \ln(T)\big) $$
$$ 3\ln(T+1) = 2\ln(T+1) + \ln(T) \tag{6} $$
Taking the exponential of $(6)$
$$ (T+1)^3 = T(T+1)^2 $$
Expanding and simplifying gives
$$ T^2 + T - 1 = 0 $$
A: After solving both differential equations for height of snow and position of snowplow as functions of time and applying the four boundary conditions, namely,
$h(t=0) = 0; x(t=T) = 0; x(t=T+1) = 2$ and $x(t=T+2) = 3$, one can algebraically combine and condense the relevant expressions down to this form:
$T(T+2)^2 = (T+1)^3$.
$T = 0.618$ or $-1.618$ hrs, assuming $t=0$ as the point in time at which it started snowing, $Δt=0-T$ must be the difference in time  between the start of snowfall and the snowplow starting to move, in that order. One may subsequently deduce that it started snowing prior to the snowplow moving, indicating that $ΔT$ must be a negative value. Therefore, $T$ must be positive, and it started snowing $37min.$ prior to the snowplow starting at $7:23 A.M.$
