When does snow usually start to fall in Big Bear, CA?

At what time did the snow start to fall?

• On a certain winter day, snow starts to fall at a heavy and steady rate. Three identical snowplows start plowing the same road, the first leaving at 12 noon, the second leaving at 1 pm, and the third leaving at 2 pm. At some time later, they all collide. At what time did the snow start to fall? The speed of a snowplow is inversely proportional to the depth of the snow.

• Answer:

  At 11:30 AM. To solve this, I made three unknown functions marking time the three snowplows pass a certain position s: t1(s), t2(s), t3(s). For uniformity, let's also define t0(s) to be a constant function of the time when the snow began to fall. The i-th snowplow passing at the point s "sees" all the snow that fell there between t_(i-1) and t_i. According to your condition, v = ds / dt_i = c/(t_i - t_(i-1)) for some constant c => dt_i / ds = (t_i - t_(i-1))/c For simplicity, let's denote b = 1/c. The initial conditions are: t0(s) = time when the snow began to fall t1(0) = 12 AM t2(0) = 1 PM t3(0) = 2 PM We can shift the time coordinate so that t0(s) = 0 for all s t1(0) = a t2(0) = a + 1 t3(0) = a + 2 Now a means for how long the snow had been falling at the noon. Let's solve this system. t1' = b*(t1 - t0) = b*t1 t1(s) = C1*exp(bs) t1(0) = C1 = a => t1(s) = a*exp(bs) t2' = b*(t2-t1) = b t2 - ab exp(bs) variation of constants: t2(s) = f(s)*exp(bs), f'(s) = -ab t2(s) = (C2 - abs) exp(bs) t2(0) = C2 = a + 1 => t2(s) = (a + 1 - abs) exp(bs) t3' = b*(t3-t2) = b t3 - (ab + b - ab^2 s) exp(bs) t3(s) = g(s)*exp(bs), g'(s) = -(ab + b - ab^2 s) t3(s) = (C3 - abs - bs + ab^2s^2/2) exp(bs) t3(0) = C3 = a + 2 => t3(s) = (a + 2 - abs - bs + ab^2s^2/2) exp(bs) Now, all of t1(s), t2(s) and t3(s) should be the same at some point s. This gives a system of equations a = a + 1 - abs = a + 2 - abs - bs + ab^2s^2/2 with unknowns a and s. We are not actually interested in s, a is the main question and all other letters are some constants defined above. The first = sign gives abs = 1, plugging this into the second one and eliminating s yields a = a + 2 - 1 - 1/a + 1/(2a) 0 = 1 - 1/(2a) 2a = 1 independently of b. Therefore, a = 1/2 hour, so the snow started to fall at 11:30 AM. Nice question! @ jason s: If "V" means my name, please note that the rate of the snowfall, the constant b, was effectively lost during the manipulations and the answer does NOT depend on it.