Chapter 2: Review

The chapter problem, solved

The ferry moves at 4.0 m s⁻¹ through the water, and the river is 120 m wide. The current flows at 3.0 m s⁻¹, at right angles to the crossing.

1. The velocity over the ground on the first crossing. The ferry points straight across, so its 4.0 m s⁻¹ is one perpendicular component and the 3.0 m s⁻¹ current is the other.

\begin{aligned} v &= \sqrt{4.0^2 + 3.0^2} \\ &= \sqrt{16 + 9} \\ &= \sqrt{25} = 5.0\,\text{m}\,\text{s}^{-1} \end{aligned}

\tan\theta = \frac{3.0}{4.0}, \qquad \theta = 36.9^\circ

The velocity over the ground is 5.0 m s⁻¹ at 36.9° to the straight-across direction, on the downstream side.

2. The crossing time. The crossing time depends on the across component and the width of the river. The current has no across component, so it does not change the time.

t = \frac{120}{4.0} = 30\,\text{s}

3. The distance downstream. The current carries the ferry downstream for the whole 30 s.

\text{distance downstream} = 3.0 \times 30 = 90\,\text{m}

The ferry lands 90 m downstream of the point directly opposite.

4. The heading to land directly opposite. To land directly opposite, the ferry must have no downstream component over the ground. Its upstream component must cancel the current, so $4.0\sin\theta = 3.0$ .

\begin{aligned} \sin\theta &= \frac{3.0}{4.0} = 0.75 \\ \theta &= \sin^{-1}(0.75) = 48.6^\circ \end{aligned}

The ferry heads 48.6° upstream of the straight-across direction.

5. The crossing time with that heading. The across component is now the cosine of the heading.

\begin{aligned} v_{\text{across}} &= 4.0\cos 48.6^\circ = 2.65\,\text{m}\,\text{s}^{-1} \\ t &= \frac{120}{2.65} = 45\,\text{s} \end{aligned}

6. The judgement. The two crossings are different trades. Pointing straight across reaches the far bank in 30 s, but the ferry lands 90 m downstream. Heading 48.6° upstream lands directly opposite, but takes 45 s, because part of the ferry speed is used to fight the current. The operator chooses between time and a straight route. This final judgement is a reasoned answer. It is not the kind of claim a computer can check, so compare your reasoning with the evidence above.

Mixed practice — choose the method first

For each question, say which rule or equation you will use before you calculate.

The ferry points straight across at $4.0\,\text{m}\,\text{s}^{-1}$ and the current is $3.0\,\text{m}\,\text{s}^{-1}$ downstream. Find the magnitude and direction of the velocity over the ground.
The river is $120\,\text{m}$ wide and the across component of the velocity is $4.0\,\text{m}\,\text{s}^{-1}$ . Find the crossing time.
The current is $3.0\,\text{m}\,\text{s}^{-1}$ and the crossing takes $30\,\text{s}$ . Find how far downstream the ferry lands.
Find the angle upstream, from the straight-across direction, at which the ferry must head so that the current is exactly cancelled.
With the heading at $48.59^\circ$ upstream, find the crossing time for the $120\,\text{m}$ river.
State the heading needed to land directly opposite, the crossing time it gives, and a reason an operator might still choose to point straight across.
Explain why the current does not change the time the ferry takes to cross the river.
Write $15\,\text{m}\,\text{s}^{-1}$ in $\text{km}\,\text{h}^{-1}$ .
Show that the equation of motion $v^2 = u^2 + 2as$ is homogeneous.
A trolley travels $25.0 \pm 0.1\,\text{m}$ in $5.0 \pm 0.2\,\text{s}$ . Find the speed and its absolute uncertainty.
A measurement gives $3.14159\,\text{m}$ with an uncertainty of $0.02\,\text{m}$ . Write the result in the correct form.
Two forces of $9.0\,\text{N}$ and $12\,\text{N}$ act at right angles. State which method you would use, and why, before you calculate the resultant.
Resolve a force of $80\,\text{N}$ at $25^\circ$ to the horizontal into horizontal and vertical components.
Two perpendicular displacements are $5.0\,\text{m}$ and $12\,\text{m}$ . Find the magnitude and direction of the resultant displacement.
A car of weight $1200\,\text{N}$ rests on a slope at $12^\circ$ to the horizontal. Find the component of the weight along the slope and the component perpendicular to it.

Answers and concise solutions

1. The velocity over the ground

Answer: Magnitude $= \sqrt{4.0^2 + 3.0^2} = 5.0\,\text{m}\,\text{s}^{-1}$ . Direction $= \tan^{-1}(3.0/4.0) = 36.9^\circ$ from the straight-across direction, towards downstream.

2. The crossing time

Answer: $t = \dfrac{120}{4.0} = 30\,\text{s}$ .

3. The distance downstream

Answer: Distance downstream $= 3.0 \times 30 = 90\,\text{m}$ . The ferry is carried $90\,\text{m}$ downstream while it crosses.

4. The heading

Answer: The upstream component must equal the current: $4.0\sin\theta = 3.0$ , so $\theta = \sin^{-1}(0.75) = 48.6^\circ$ upstream of the straight-across direction.

5. The crossing time with the heading upstream

Answer: The across component is $4.0\cos 48.6^\circ = 2.65\,\text{m}\,\text{s}^{-1}$ , so $t = \dfrac{120}{2.65} = 45\,\text{s}$ . The crossing is longer because part of the ferry speed is used to fight the current.

6. The judgement

Head about 49 degrees upstream of the straight-across direction, so the upstream component cancels the current.
The crossing then takes about 45 s, longer than the 30 s straight-across crossing, because the across component falls to 2.65 m/s.
Pointing straight across reaches the far bank in 30 s, but the ferry lands 90 m downstream, so the operator might accept the drift to save time.

Answer: Head about $49^\circ$ upstream to land directly opposite; the crossing then takes about $45\,\text{s}$ . Pointing straight across is quicker at $30\,\text{s}$ , but the ferry lands $90\,\text{m}$ downstream.

7. Why the current does not change the crossing time

The two velocities are at right angles, so the current has no component along the straight-across direction.
The crossing time depends only on the across component of the velocity and the width of the river. The current changes the direction over the ground and the distance downstream, but not the across component.

Answer: The current is at right angles to the crossing, so it has no component across the river. The crossing time depends only on the across component and the width, so the current does not change it.

8. The speed in kilometres per hour

Answer: $15\,\text{m}\,\text{s}^{-1} = 15 \times \dfrac{3600}{1000} = 54\,\text{km}\,\text{h}^{-1}$ .

9. A homogeneity check

Answer: The left side has base units $(\text{m}\,\text{s}^{-1})^2 = \text{m}^2\,\text{s}^{-2}$ . The right side has $u^2 = \text{m}^2\,\text{s}^{-2}$ and $as = \text{m}\,\text{s}^{-2} \times \text{m} = \text{m}^2\,\text{s}^{-2}$ . All three terms match, so the equation is homogeneous.

10. The speed and its uncertainty

Answer: The percentage uncertainty in the distance is $\dfrac{0.1}{25.0} \times 100\% = 0.4\%$ , and in the time it is $\dfrac{0.2}{5.0} \times 100\% = 4.0\%$ . For a quotient, add them: $0.4\% + 4.0\% = 4.4\%$ . The speed is $\dfrac{25.0}{5.0} = 5.0\,\text{m}\,\text{s}^{-1}$ , so its absolute uncertainty is $5.0 \times 0.044 = 0.22\,\text{m}\,\text{s}^{-1}$ . The speed is $5.0 \pm 0.2\,\text{m}\,\text{s}^{-1}$ .

11. The result in the correct form

Answer: The uncertainty is $0.02\,\text{m}$ to one significant figure, so the value is quoted to the same decimal place: $3.14 \pm 0.02\,\text{m}$ .

12. The method and its reason

The two forces are at right angles, so they and their resultant form a right-angled triangle.
Find the magnitude with Pythagoras, and the direction from the tangent of the angle. This is quicker and more accurate than a scale drawing because the vectors are perpendicular.

Answer: The forces are at right angles, so use Pythagoras: the resultant has magnitude $\sqrt{9.0^2 + 12^2} = 15\,\text{N}$ and direction $\tan^{-1}(12/9.0) = 53.1^\circ$ from the $9.0\,\text{N}$ force.

13. The two components

Answer: Horizontal component $= 80\cos 25^\circ = 72.5\,\text{N}$ . Vertical component $= 80\sin 25^\circ = 33.8\,\text{N}$ .

14. The resultant displacement

Answer: Magnitude $= \sqrt{5.0^2 + 12^2} = 13\,\text{m}$ . Direction $= \tan^{-1}(12/5.0) = 67.4^\circ$ from the $5.0\,\text{m}$ displacement.

15. The two components of the weight

Answer: Along the slope: $1200\sin 12^\circ = 249.5\,\text{N}$ . Perpendicular to the slope: $1200\cos 12^\circ = 1173.8\,\text{N}$ . The perpendicular component is much larger because the slope is shallow.

What you can now do

tell a scalar quantity from a vector quantity;
explain why a stated direction alone does not make a quantity a vector;
write a vector as a magnitude and a direction;
use positive and negative signs to give the direction of a vector along a line;
resolve a vector into two perpendicular components with the sine and cosine;
choose the cosine component from the direction the angle is measured from;
add two perpendicular vectors with Pythagoras and find the direction with the tangent;
resolve a weight into components along and perpendicular to a slope;
describe how a scale drawing adds vectors that are not at right angles;
use vectors to solve a navigation problem.

Your relevance reflection

Return to the prompt from the chapter opener: think of a journey you make often, and compare the distance travelled with the straight-line displacement. Write one more example now that you have finished the chapter. Has the chapter changed how you think about a route, a map or a stated speed?