Question

A freight train travels at 30 mph. If the freight train is 300 miles ahead of an express train traveling at 55 mph, how long will the express train take to overtake the freight train? I at least need to know how to set up the problem.

Answer 1

Since each speed is constant, the position of each train is given by a linear equation. From the definition of speed, \[(speed) = \frac{(distance)}{(time)}\rightarrow (distance)=(speed)(time)\]That is,\[x_1=v_1t+x_{0,1}\]\[x_2=v_2t+x_{0,2}\]where x_1 is the express train and x_2 is the freight. The terms\[x_{0,1},x_{0,2}\]are the initial displacements of your trains (that is, where they are at time t=0).
We know that, at time 0, the different between their distances is 300 miles; that is,\[x_2-x_1=(v_2-v_1)t+(x_{2,0}-x_{1,0}) \rightarrow 300=x_{2,0}-x_{1,0}\]So the equations become,\[x_1=v_1t+x_{0,1}\]\[x_2=v_2t+x_{1,0}+300\]You want to find the time where both trains are at the same point; that is, when they've traveled the same distance, so when

Answer 2

\[x_1=x_2\]that is, when\[v_1t+x_{1,0}=v_2t+x_{1,0}+300 \rightarrow v_1t=v_2t+300\]\[\rightarrow t=\frac{300}{v_1-v_2}=\frac{300}{55-30}=12hrs\]

Answer 3

You could also consider it geometrically in the (t,x) coordinate plane (i.e. make t your 'x-axis' and x your 'y-axis'). Then the slope is (change in distance)/(change in time) which is the speed.

At t=0, the freight train will be at x=300 and the express will be at x = 0.

Draw the lines, and where they intersect is the point where they simultaneously share both the same distance at the same time.