Problem1C-01*: Motion along a straight line

The motion of a particle in a flat plane is described by its x and y coordinates at time t, that is, by x(t) and y(t). Suppose

x(t)=At+B    and    y(t)=Ct+D.                                        (1)

(a) What is the length of its position vector expressed in terms of t?

(b) If expressed in polar coordinate form, what is the angle between the position vector and the x-axis at t = 0? How about
at t = t1 if B = D = 0?

(c) Describe the trajectory of the motion of the particle.

Scroll down for solution

Solution:

(a) Let the length of the position vector of the particle be denoted as r(t). Since

``` ```
by substituting Eqs.(1) into the above expression, we have
``` ```

(b) In polar coordinate representation as shown in the figure at the right, we have

x(t) = r(t)cos{θ(t)},
(3)
y(t) = r(t)sin{θ(t)}.

At t = 0, we have from Eqs.(1) and (2)

``` ```
Thus Eq.(2) with t = 0 becomes
``` ```
We should note that in Eq.(4) both cos-1 amd sin-1 are necessary to determine θ(0) without ambiguity.

At t = t1 with B = D = 0, Eqs.(1) and (2) become

x(t1) = At1,    y(t1) = Ct1
and
``` ```
Again in Eq.(5) both cos-1 amd sin-1 are necessary to determine θ(t1) completely.

(c) Eqs.(1) can be transformed to At = x-B and Ct = y-D. We need to eliminate t from those two equations to get an equation relating x and y. We need to consider the following possibilities:

(i) If A = C = 0, then x = B and y = D. Thus the particle stays at the point (B, D) and the trajectory is just a point.

(ii) If A = 0 but C ≠ 0, then x = B and y = Ct+D. The trajectory is a straight line parallel to the y-axis and intercept the x-axis at x = B. At t = 0 the particle is located at the point (B, D), then the particle moves up or down the line depending on the sign of C.

(iii) If A ≠ 0 and C = 0, then x = At+B and y = d. The trajectory is a straight line parallel to the x-axis and intercept the y-axis at y = D. At t = 0 the particle is at the point (B, D) and moves along the line toward the right or the left accoeding to the sign of A.

(iv) If A ≠ 0 and C ≠ 0, then

``` ```
Elliminating t, we get (x-B)/A = (y-D)/C or
``` ```
This equation represents a trajectory that is a straight line with a slope C/A, and intercepts the y-axis at y = (D/C - B/A).