Motion in a Plane

From the chapter, scalars have magnitude only, while vectors have both magnitude and direction.

- Volume — scalar
- Mass — scalar
- Speed — scalar
- Acceleration — vector
- Density — scalar
- Number of moles — scalar
- Velocity — vector
- Angular frequency — scalar
- Displacement — vector
- Angular velocity — vector

From the chapter, the two scalar quantities in the list are work and current. The others listed are vectors or are treated as vectors in this context.

The only vector quantity in the list is impulse. The other quantities given are scalars.

- (a) Adding any two scalars — meaningful, provided they are of the same type/unit.
- (b) Adding a scalar to a vector of the same dimensions — not meaningful; scalars and vectors are different kinds of physical quantities.
- (c) Multiplying any vector by any scalar — meaningful; the result is another vector.
- (d) Multiplying any two scalars — meaningful; the result is a scalar.
- (e) Adding any two vectors — meaningful; vectors obey vector addition laws.
- (f) Adding a component of a vector to the same vector — not meaningful; a component is a scalar, while the vector itself is a vector.

Using the chapter:

- (a) True — magnitude of a vector is a scalar.
- (b) True — each component of a vector is a scalar.
- (c) False — total path length is generally greater than or equal to displacement magnitude; they are equal only in straight-line motion without change of direction.
- (d) True — average speed is greater than or equal to the magnitude of average velocity.
- (e) False — three non-coplanar vectors can add to give a null vector; for example, vectors can close in 3D to form a closed triangle/polygon.

For two vectors a and b, the magnitude of their sum is always between the sum and the difference of their magnitudes.

- (a) By the triangle law, the resultant side of a triangle cannot be longer than the sum of the other two sides, so
$$|\mathbf{a}+\mathbf{b}| \le |\mathbf{a}|+|\mathbf{b}|.$$

- (b) Also, the resultant must be at least the difference of the two side lengths, so
$$|\mathbf{a}+\mathbf{b}| \ge ||\mathbf{a}|-|\mathbf{b}||.$$

These are the standard triangle inequalities for vectors.

The initial and final points are diametrically opposite on a circle of radius $200\,\text{m}$. So the displacement is the straight line joining them, i.e. the diameter:

$$|\Delta \mathbf{r}| = 2R = 2 \times 200 = 400\,\text{m}.$$

This displacement is equal to the actual path length only for the girl who moves along the straight-line path PQ.

The cyclist starts from the centre and returns to the centre.

- (a) Net displacement: initial and final positions are the same, so displacement is zero.
$$\Delta \mathbf{r}=0$$

- (b) Average velocity:
$$\overline{\mathbf{v}}=\frac{\Delta \mathbf{r}}{\Delta t}=\frac{0}{10\,\text{min}}=0$$

- (c) Average speed: total distance = radius to edge $1\,\text{km}$ + semicircular/circular part as shown in the textbook figure + return radius $1\,\text{km}$. Since the route is $1$ km out, half circumference of radius $1$ km along the edge from $P$ to $Q$, then $1$ km back, the total distance is
$$1 + \pi(1) + 1 = (2+\pi)\,\text{km}$$
Hence
$$\text{average speed} = \frac{2+\pi}{10}\,\text{km/min} \approx 0.514\,\text{km/min}$$

However, the textbook exercise figure indicates the cyclist goes from O to P, then along the circumference to Q, then back to O; so using the full intended path, the average speed is based on that total distance over 10 min.

Each turn is after 500 m, and the direction changes by 60° to the left each time.

- At the third turn: path length covered = $3\times 500 = 1500\,\text{m}$.
The vector sum of three successive sides of a hexagonal-type path gives a resultant displacement of 1500 m.

- At the sixth turn: path length = $6\times 500 = 3000\,\text{m}$.
After six 60° turns, the motorist completes a closed hexagonal pattern, so displacement is 0 if the path returns to the start.

- At the eighth turn: path length = $8\times 500 = 4000\,\text{m}$.
The displacement is the resultant of the first six segments plus two more, so it is not equal to the path length and is smaller.

Thus, in every case the magnitude of displacement is less than or equal to path length; equality would hold only if motion were along a straight line.

Given:
- Total distance travelled = 23 km
- Straight-line displacement = 10 km
- Time taken = 28 min = 28/60 h

(a) Average speed
$$\text{Average speed} = \frac{\text{total path length}}{\text{time}} = \frac{23}{28/60}$$
$$= 23\times \frac{60}{28} = 49.29 \approx 49.3\,\text{km/h}$$

(b) Magnitude of average velocity
$$|\overline{\mathbf{v}}| = \frac{\text{displacement}}{\text{time}} = \frac{10}{28/60}$$
$$= 10\times \frac{60}{28} = 21.43 \approx 21.4\,\text{km/h}$$

They are not equal because the path length is greater than the displacement.

This is the textbook maximum horizontal distance result.

For a ball thrown with speed $40\,\text{m s}^{-1}$ from a hall with ceiling height $25\,\text{m}$, the maximum horizontal distance without hitting the ceiling is 50 m.