Average Velocity and Instantaneous Velocity — Question 2

Question 2

Unequal intervals require time weighting

A particle moves with constant velocity $(4,0)$ m/s for 3 s, then $(-2,3)$ m/s for 6 s. Its initial position is the origin.

1. Find both leg displacements, the final position and the full-trip average velocity.

2. Compare with the arithmetic mean of the two velocity vectors.

3. Find the average speed and state when the simple arithmetic mean of segment velocities would be justified.

Question 2 - Solution

Unequal intervals require time weighting

(a) Convert each velocity into its own displacement.

$$\Delta\vec r_1=3(4,0)=(12,0),\quad\Delta\vec r_2=6(-2,3)=(-12,18),\quad\boxed{\vec v_{\mathrm{av}}=(0,2)\,\mathrm{m/s}}$$ The legs give $(12,0)$ m and $(-12,18)$ m. Final position is $(0,18)$ m, so $\vec v_{\mathrm{av}}=(0,2)$ m/s over 9 s. The horizontal displacements cancel despite unequal horizontal speeds.

(b) Weight by elapsed time.

$$\frac{\vec v_1+\vec v_2}{2}=(1,1.5)\ne\frac{3\vec v_1+6\vec v_2}{9}=(0,2)\quad\mathrm{m/s}$$ The unweighted mean is $(1,1.5)$ m/s, which is wrong here. The correct expression is $(3\vec v_1+6\vec v_2)/9$. A velocity maintained twice as long contributes twice its weight to the interval average.

(c) Average the traveled lengths.

$$\bar v_{\mathrm{speed}}=\frac{3(4)+6\sqrt{13}}9=3.737\,\mathrm{m/s},\quad\Delta t_1=\Delta t_2\ \Rightarrow\ \vec v_{\mathrm{av}}=\tfrac12(\vec v_1+\vec v_2)$$ Average speed is $(12+6\sqrt{13})/9\approx3.737$ m/s. The unweighted mean of two constant segment velocities is justified for equal durations (or when a special equality makes the weighting irrelevant), not simply because there are two segments.