Q.
Suppose you are late to catch a bus. You are 10 m away from the bus stop you see it pull away. It accelerates from rest at 0.9 m/s2. Assuming you can run at a constant speed, what is the minimum speed with which you can run toward the bus and still catch it (assuming you're into that "Indiana Jones" grab-onto-the-back-of-the-moving-vehicle technique)?

On the surface, this doesn't seem to be too difficult to deal with. Using the kinematic equations to track both your position and the bus', we realize that when you reach the bus, your position is the same as that of the bus.

But first, let's begin by jotting down everything we know about the situation. Let's declare the origin (x = 0) to be where you begin to run.


Next the equations of motion for you and the bus:

... and finally, when you finally reach the bus, these position equations are equal to each other:

Hmmm.... Seems as if my intuition didn't carry us far enough. We've got only one equation, but there are two unknowns.

Well, there are several ways to approach this question, but I feel that the following discussion is most enlightening. It involves the use of graphs. Let's take a look at the following position vs. time graph.


The equation for the bus is plotted exactly according to what we have above. Notice how it begins at a position of 10 at t=0 and curves upwards like a parabola because of the t2.

The straight line, which represents your position, is plotted supposing a speed of 5 m/s. Mind you, this is not necessarily your speed - it's just an example.

Recall, that for a plot showing the motion of an object moving at a constant velocity, the slope of the graph is the speed. The more vertical the line - the greater the speed.

For an object that accelerates, the plot will be parabolic. The speed at any point is the slope of the straight line that is tangent to the curve at that point. Confused? Well, we'll come back to this later.


The black dots in the above graph show the points where your position is the same as that of the bus. In this case, it looks as if the positions are equal at two instances in time. How can this be?
It can happen, if at the first instance you pass the bus since your speed was greater. But the bus is accelerating, gaining speed, so eventually it overtakes you and your positions are equal once again as the bus passes you, this time for good.

Well, we want the minimum speed with which you can catch the bus. In this case, 5 m/s was too fast. We want the speed to be such that your positions meet just once, and just barely at that.


If your speed was any slower, the slope of your line would be lower, and your position would never be equal to that of the bus.

Since your straight line touches the curved line of the bus at just one point, it is therefore tangent to that curve. This means that at the point where the positions are equal, the slopes of the two curves are equal - and this leads us to a second equation. The minimum speed will be such that


where t is the time when the intersection occurs.
To find t we need to apply a bit of simple algebra, along with the equation we found before...

This is how long it takes for you to catch the bus. Now, to find your minimum speed:



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