Physics

Physics

Problem

At t=0t=0, two particles have position vectors (in metres) rA0=2i^+3j^\mathbf r_{A0}=2\hat{i}+3\hat{j} and rB0=8i^−5j^\mathbf r_{B0}=8\hat{i}-5\hat{j}. Their constant velocity vectors are vA=4i^+0j^\mathbf v_A=4\hat{i}+0\hat{j} and vB=−2i^+6j^\mathbf v_B=-2\hat{i}+6\hat{j} respectively. a. (i) Write the position vectors of particles A and B as functions of time tt. b. (ii) Hence obtain the position vector of A relative to B and the square of their separation as functions of time. c. (iii) Find the time at which the two particles are closest to each other. d. (iv) Determine this minimum separation.
Intuitive explanation

The key is to write each particle's position as a function of time using r(t)=r0+vt\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}t, then use the constraint that they collide when their positions are equal. This reduces a 2D kinematics problem to solving a single equation for tt—the relative motion perspective makes it clear whether (and when) the particles will meet.

Intuitive explanation

The key is to write each particle's position as a function of time using r(t)=r0+vt\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}t, then use the constraint that they collide when their positions are equal. This reduces a 2D kinematics problem to solving a single equation for tt—the relative motion perspective makes it clear whether (and when) the particles will meet.

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