Adding Vectors Geometrically — Question 7

Question 7

Subtracting a vector means adding its reversed arrow

Two velocity vectors are drawn tail-to-tail at O in one laboratory reference. $\vec u$ is 4 m/s east with head U, and $\vec v$ is 3 m/s north with head V. Define $\vec d=\vec v-\vec u$ as the vector satisfying $\vec u+\vec d=\vec v$. No time interval is supplied.

1. Identify the directed arrow representing $\vec d$ in the tail-to-tail diagram. Find its magnitude and direction by right-triangle geometry.

2. Construct the same vector using $\vec v+(-\vec u)$ head-to-tail. What must be preserved when drawing $-\vec u$?

3. Compare $\vec v-\vec u$ with $\vec u-\vec v$. Diagnose the proposed answer $|\vec d|=3-4=-1\ \mathrm{m/s}$.

4. If these are an object’s initial velocity $\vec u$ and final velocity $\vec v$, what does $\vec d$ represent? Does it determine average acceleration or distance traveled without further information?

Question 7 - Solution

Subtracting a vector means adding its reversed arrow

(a) Find the arrow that completes O–U–V.

$$\vec d=\overrightarrow{UV}=(-4,3)\,\mathrm{m/s},\quad\boxed{d=5\,\mathrm{m/s}},\quad\alpha=\tan^{-1}(3/4)=36.870^\circ$$ The difference runs from the head U of the subtracted vector to the head V of the first vector. Its direction is $36.870^\circ$ north of west.

(b) Reverse direction while keeping length.

$$\vec v+(-\vec u)=(0,3)+(-4,0)=(-4,3)\,\mathrm{m/s}$$ Reverse u while preserving its 4 m/s magnitude, then translate it to v’s head. The resulting directed separation is the same difference.

(c) Reversing subtraction reverses the difference.

$$\vec u-\vec v=-\vec d,\quad|-\vec d|=5\,\mathrm{m/s}\ne3-4\,\mathrm{m/s}$$ Changing subtraction order reverses direction without changing magnitude. Subtracting magnitudes ignores the perpendicular geometry; even its absolute value, 1 m/s, would be incorrect.

(d) Endpoint velocities determine their change, not the whole motion.

$$\Delta\vec v=\vec d,\qquad\bar{\vec a}=\frac{\Delta\vec v}{\Delta t},\qquad s=\int_{t_i}^{t_f}|\vec v(t)|\,dt$$ The difference specifies velocity change. Average acceleration needs elapsed time, and distance needs the speed history; neither follows from the two endpoint velocities alone.