Interactive Lecture
Understanding Motion from Different Perspectives
by Dr Arnaldo Delli Carri
Motion is always observed from a particular frame of reference. The same event — a car overtaking, a ball thrown on a moving ship — looks completely different depending on where you're watching from. You use exactly this difference every time you set up a relative velocity or relative acceleration problem.
All motion is relative. There is no such thing as absolute motion – motion can only be described relative to a chosen reference frame.
Definition: The relative displacement of object B with respect to object A is the vector difference between their position vectors:
\(\vec{r}_{B/A} = \vec{r}_B - \vec{r}_A\)
Where:
In 2D Cartesian coordinates:
\(\vec{r}_{B/A} = (x_B - x_A)\hat{i} + (y_B - y_A)\hat{j}\)
Magnitude: \(|\vec{r}_{B/A}| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}\)
Direction: \(\theta = \tan^{-1}\left(\frac{y_B - y_A}{x_B - x_A}\right)\)
\(\vec{r}_{B/A} = -\vec{r}_{A/B}\)
The position of B relative to A is the negative of the position of A relative to B.
Definition: The relative velocity of object B with respect to object A is the vector difference between their velocity vectors:
\(\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A\)
Where:
For three reference frames (e.g., A, B, and ground G):
\(\vec{v}_{B/G} = \vec{v}_{B/A} + \vec{v}_{A/G}\)
This states that the velocity of B relative to the ground equals the velocity of B relative to A plus the velocity of A relative to the ground.
In 2D Cartesian coordinates:
\(\vec{v}_{B/A} = (v_{Bx} - v_{Ax})\hat{i} + (v_{By} - v_{Ay})\hat{j}\)
Speed: \(|\vec{v}_{B/A}| = \sqrt{(v_{Bx} - v_{Ax})^2 + (v_{By} - v_{Ay})^2}\)
Explore relative motion by observing two objects from different reference frames. Switch between viewpoints to see how motion appears different from each perspective.
A: 2.0 m/s @ 45°
B: 3.0 m/s @ 135°
| Concept | Equation | Description |
|---|---|---|
| Relative Position | \(\vec{r}_{B/A} = \vec{r}_B - \vec{r}_A\) | Position of B relative to A |
| Relative Velocity | \(\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A\) | Velocity of B relative to A |
| Reciprocity | \(\vec{v}_{B/A} = -\vec{v}_{A/B}\) | Relative velocities are opposite |
| Velocity Addition | \(\vec{v}_{C/A} = \vec{v}_{C/B} + \vec{v}_{B/A}\) | Chain rule for reference frames |
| Component Form | \(\vec{v}_{B/A} = (v_{Bx} - v_{Ax})\hat{i} + (v_{By} - v_{Ay})\hat{j}\) | 2D relative velocity components |
A boat can travel at 5 m/s in still water. It needs to cross a river that flows east at 3 m/s. If the river is 100 m wide:
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Solution:
Write each velocity in frame-relative form: the boat relative to the water \(v_{b/w} = 5\) m/s, the water relative to the ground \(v_{w/g} = 3\) m/s (flowing east), and the boat relative to the ground \(v_{b/g}\).
1) To travel straight across, the upstream component of the boat's heading must cancel the current:
\(v_{b/w} \sin\theta = v_{w/g} = 3\) m/s
\(\sin\theta = \frac{3}{5} = 0.6 \implies \theta = 36.87°\) west of north
2) Speed over the ground is the surviving northward component:
\(v_{b/g} = v_{b/w} \cos\theta = 5 \times 0.8 = 4\) m/s
3) Time to cross the 100 m width:
\(t = \frac{d}{v_{b/g}} = \frac{100}{4} = 25\) s
Answer: head 36.87° west of north; ground speed 4 m/s; crossing time 25 s.
An aircraft flies at 200 m/s relative to the air. There's a wind blowing from the west at 50 m/s. The pilot wants to fly northeast (45° from north). What heading should the pilot take, and what will be the ground speed?
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Solution:
Velocity addition: \(\vec{v}_{p/g} = \vec{v}_{p/a} + \vec{v}_{a/g}\), where \(v_{p/a} = 200\) m/s is the plane relative to the air and \(v_{a/g} = 50\) m/s is the wind — the air relative to the ground, blowing toward the east.
Let the heading be \(\theta\) east of north, with east as \(\hat{i}\) and north as \(\hat{j}\):
\(200\sin\theta + 50 = v_{p/g} \sin(45°)\) …(1)
\(200\cos\theta + 0 = v_{p/g} \cos(45°)\) …(2)
Dividing (1) by (2) and noting that \(\sin 45° = \cos 45°\):
\[200\sin\theta + 50 = 200\cos\theta \implies \cos\theta - \sin\theta = 0.25\]
Using the identity \(\cos\theta - \sin\theta = \sqrt{2}\cos(\theta + 45°)\):
\[\cos(\theta + 45°) = \frac{0.25}{\sqrt{2}} = 0.1768 \implies \theta + 45° = 79.82° \implies \theta = 34.82°\]
The pilot heads slightly less east than the desired 45° track, to account for the eastward wind push.
Ground speed from equation (2):
\[v_{p/g} = \frac{200\cos(34.82°)}{\cos(45°)} = \frac{200 \times 0.8210}{0.7071} = 232.2 \text{ m/s}\]
Answer: head 34.82° east of north; ground speed 232.2 m/s.
Two cars approach an intersection. Car A travels north at 15 m/s, whilst Car B travels east at 20 m/s. What is the velocity of Car B relative to Car A? At what angle does Car A's driver see Car B approaching?
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Solution:
Relative velocity of B with respect to A:
\(\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A\)
\(\vec{v}_{B/A} = 20\hat{i} - 15\hat{j}\) m/s
Magnitude:
\(|\vec{v}_{B/A}| = \sqrt{20^2 + 15^2} = \sqrt{625} = 25\) m/s
Direction (from Car A's perspective):
\(\theta = \tan^{-1}\left(\frac{20}{15}\right) = 53.13°\) east of south
Both cars are closing on the intersection, so B sits to the northwest of A. Its relative velocity points 53.13° east of south — toward the southeast — so Car A's driver sees Car B approaching from the northwest at 25 m/s.
Answer: \(\vec{v}_{B/A} = 20\hat{i} - 15\hat{j}\) m/s, magnitude 25 m/s at 53.13° east of south; B approaches from the northwest.