Answer to the August 15, 1997 Question

Question

Physics 211 - Suppose that a particle moves with constant velocity v = -15 m/s along an x-axis and that it passes the point x = 20 m at t = - 1 s. Where will the particle be at t = 3 s? Solve this problem both analytically, i.e., by calculation, and graphically.

Also derive an expression for the general position x of the particle at the general time t. In addition, make a diagram that shows the x-axis and the particle's positions on the x-axis at t = -1 s and t = 3 s. (Definition of velocity. Analytical geometry. Graphs.)

Answer

The velocity v is the time-rate-of-change of position, which means that it is equal to the ratio (change in position)/(change in time).

With x(3) denoting the position at t = 3 s and x(-1) the position at t = -1 s, we can express this ratio in terms of the given and unknown quantities as

v = [(x(3) - x(-1)]/[3 - (-1)] = [x(3) - 20]/4 = - 15.(1)

Solving this equation for x(3), gives

x(3) = 20 - 15x4 = -40 m.

The graphical solution is as follows. We plot the given point (t = -1, x = 20) and draw a line through this point that has a slope equal to v = -15. To construct the slope, we use the given value of v (which is equal to the slope, to construct a second point. Since v = -15, we know that when t increases by 1, then x changes by -15, or when t increases by 2, then x changes by 2x(-15) = -30. Thus, as t changes by 2 from -1 to 1, x changes from 20 to 20 + (-30) = -10. The point (t = 1, x = -10) is plotted in the graph below. Then we draw a straight line through the two points.

From this graph we can read off the value of x for any value of t. At t = 3 s, the graph gives us x = -40 m. See the following graph.

Note that the calculation we had to do to calculate the second point in order to be able to draw the graph is tantamount to setting up equation (1). Thus, the graphical solution is not really an alternate way to finding x(3). However, once we have the graph, it allows us to quickly find the value of x at any t by simply looking at the graph.

To find a general expression for x at a general t, all we have to do is to replace x(3) by x and the "3" in the denominator in equations (1) by t. This results in the equation

[x - 20]/[t - (-1)] = -15.

Solving for x, we obtain

x = 20 - 15(t + 1) = 20 - 15t - 15 = 5 - 15t.(2)

Note that you find x = -40 m when you substitute t = 3 s into expression (2).

Finally, our results can be summarized as in the following diagram

Compare also the May 16, 1997 question.


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