Relative Motion in Two Dimensions — Question 4

Question 4

Closest approach need not be a collision

At t=0, A is at $(0,0)$ m and B at $(100,60)$ m. Their constant velocities are $(2,1)$ and $(-2,-2)$ m/s, respectively. Treat them as points.

1. Write the position of B relative to A and test whether they collide.

2. Find the time of closest approach for $t\geq0$.

3. Find the minimum separation and verify that relative position is perpendicular to relative velocity then.

Question 4 - Solution

Closest approach need not be a collision

(a) Subtract simultaneous positions.

$$\vec r_{B/A}=(100,60)+(-4,-3)t,\quad t_x=100/4=25\,\mathrm s\ne t_y=60/3=20\,\mathrm s$$ $\vec r_{B/A}=(100-4t,60-3t)$ m for t in seconds. Zero x separation requires t=25 s, while zero y separation requires t=20 s. These conditions do not occur together, so no collision occurs.

(b) Minimize squared distance.

$$\frac{dD^2}{dt}=2\vec r_{B/A}\cdot\vec v_{B/A}=0\ \Rightarrow\ \boxed{t_*=-\frac{\vec r_0\cdot\vec v}{v^2}=\frac{580}{25}=23.2\,\mathrm s}$$ $D^2=(100-4t)^2+(60-3t)^2$. Its derivative vanishes when $-4(100-4t)-3(60-3t)=0$, giving $25t=580$ or t=23.2 s. The positive quadratic coefficient makes this the minimum, and the time lies in the allowed domain.

(c) Evaluate and check the geometry.

$$\vec r_*=(7.2,-9.6)\,\mathrm m,\quad D_{\min}=12\,\mathrm m,\quad\vec r_*\cdot\vec v=-28.8+28.8=0$$ At that time the relative position is $(7.2,-9.6)$ m, so $D=12$ m. Its dot product with relative velocity $(-4,-3)$ m/s is $-28.8+28.8=0$. The closest point on the relative straight-line path is reached when further motion is perpendicular to the separation.