Average Velocity and Instantaneous Velocity — Question 8

Question 8

Constant velocity fixes a directed straight trajectory

A particle starts at $(1,-2,3)$ m and maintains constant velocity $(2,-1,2)$ m/s for 5 s.

1. Write its position vector as a function of time and find its speed.

2. Does it pass through $(7,-5,9)$ m? If so, when?

3. A waypoint $(7,-5,8)$ m has two matching coordinates at that time. Is it on the trajectory? Explain the consistency requirement.

Question 8 - Solution

Constant velocity fixes a directed straight trajectory

(a) Integrate the constant vector rate.

$$\vec r=\vec r_0+\vec vt,\quad\boxed{v=\sqrt{2^2+(-1)^2+2^2}=3\,\mathrm{m/s}}$$ $\vec r(t)=(1,-2,3)\ \mathrm m+(2,-1,2)\ \mathrm{m/s}\,t$. The speed is $\sqrt{4+1+4}=3$ m/s. This is a straight line traversed at a constant rate, not merely a statement about endpoint displacement.

(b) Require one common time.

$$t_x=\frac{7-1}{2}=3,\quad t_y=\frac{-5+2}{-1}=3,\quad t_z=\frac{9-3}{2}=3\quad\mathrm s$$ The x equation gives $t=(7-1)/2=3$ s. At 3 s, y equals $-5$ m and z equals 9 m, so all three conditions agree. The point is reached during the specified 5 s interval.

(c) Check the remaining coordinate.

$$t_z'=\frac{8-3}{2}=2.5\,\mathrm s\ne t_x=t_y=3\,\mathrm s$$ The second waypoint would require $t=(8-3)/2=2.5$ s from z, conflicting with the 3 s required by x and y. It is not on this directed trajectory. Matching a projection does not establish a three-dimensional encounter.