**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 = t_{1} if
B = D = 0?

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

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**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

At t = t

x(t

and

Again in Eq.(5) both cos

(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).

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