First-Order Differential Equations
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Two Snowplows - Differential Equations (First-Order Differential Equations)
One day it began to snow exactly at noon at a heavy and steady rate. A snowplow left its garage at 1:00pm, and another one followed in its tracks at 2:00pm.
a)At what time did the second snowplow crash into the first? To answer this question, assume that the rate (in mph) at which a snowplow can clear the road is inversely proportional to the depth of the snow (and hence to the time elapsed since the road was clear of snow). [Hint: begin by writing differential equations for x(t) and y(t), the distance traveled by the first and second snowplows, respectively, at t hours past noon. To solve the differential equation involving y, let t rather than y be the dependent variable]
b)Could the crash have been avoided by dispatching the second snowplow at 3:00pm instead?
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Differential Equations are featured. The expert examines the first-order differential equations.
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Explanations
In addition to the notations in Figure 2.14, let us denote t_1=1h and t_2=2h the start times. That is the 1st plow starts at
t=t_1 , x(t_1 )=0 (i1)
and the 2nd plow starts at
t=t_2 , y(t_2 )=0 (i2)
These are the initial conditions for the differential equations (DE) we shall derive for x(t) and y(t).
The speed of a plow is inversely proportional to the height of the snow cover, and this height is in turn proportional to the time passed since the moment when snow cover started to grow from zero.
In the case of the 1st plow, this time is just (t-0)=t, so we take the speed of the 1st plow to be
dx(t)/dt=k/t, (1)
where k is a constant coefficient determined by the rate of the snowfall and the rate at which the plow removes snow. It numerical value will not be needed in the discussion. To solve equation (1) we can just integrate it over dt from t=t_1 to any t:
x(t)-x(t_1 ...
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