This applet helps you better understand the link between the visual and graphical approaches to the time, rate,
distance problem and its algebraic solution.
1-learn how to use the applet to approximate a solution to the problem.
2-find an algebraic solution to the problem with the help of the applet.
At 11:00am, a car (1) leaves city "A" at a constant rate of 60 mi/hr toward city "B". At the same time a second car (2) leaves city "B" toward city "A" at the constant rate of 50 mi/hr. The distance between cities A and B is 220 miles and these cities are connected by a highway used by the two cars.
At what time will the two cars cross each other?
1 - click on the button above " click here to start " to start the applet. Note the simplified picture of the two cities and the lines of the highway (blue and red).
2 - check that the speeds are 60 for car(1), 50 for car(2) and the distance is 220.
3 - click on the button "continuous".The two cars leave cities A and B. On the right side, two graphs of the distances of cars(1) (in blue) and (2) (in red) from city A are shown. The graph in blue starts from a distance equal to zero while the graph in red starts from a distance equal to 220 miles. When the two cars meet, the two graphs intersect.
4 - click on "pause", the animation is suspended. Click on pause again,the animation continues.
5 - click on "resetall" as this will reset the cars at their initial positions.
6 - to be able to answer the question above, you need to click
on "by_step" continuously till the two graphs intersect. Read the time in minutes given in the left hand corner. This is a good approximation to the answer of the question in problem 1.
distance d1, between city "A" and car (1), changes with the time t as
d1 = 60t , t = 0 corresponds to 11:00am.
distance d2, between city "A" and car (2),
changes with the time t as
d2 = 220 - 50t
when the cars cross each other d1 = d2 (blue and red graphs intersect).
60t = 220 - 50t
110t = 220
t = 2 hours
the two cars cross each other at
11:00am + 2hours = 13:00pm
Compare to the approximation obtained graphically above.