AshPhys

Combining velocities

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.1Chapter 1: KinematicsLesson 1.6Sections 1.6

Goal: Add velocities as vectors, for boats crossing rivers and aircraft in a wind.

velocity

01The big picture

When something moves through a medium that is itself moving (a boat in a current, a plane in a wind), its velocity relative to the ground is the vector sum of the two velocities.

The parts act independently: the time to cross a river depends only on the velocity across it, and the current only carries the boat downstream while it crosses.

Resultant velocity [paraphrase]
The vector sum of the velocities: the actual velocity relative to the ground.
Fig. 1 · A boat crossing a river
A boat crossing a riverTop view of a river between two banks. From the near bank, the boat velocity v_b points straight across, the current v_c points downstream from its tip, and the resultant v runs diagonally from the start to the end of v_c, at angle theta to straight across.vbvcvθcurrent
The boat heads straight across (vbv_\mathrm{b}); the current (vcv_\mathrm{c}) carries it downstream, so it actually moves along vv.

02Key equations

Resultant (perpendicular)
v=vb2+vc2v = \sqrt{v_\mathrm{b}^{2} + v_\mathrm{c}^{2}}
Angle from straight across
tan⁡θ=vcvb\tan\theta = \frac{v_\mathrm{c}}{v_\mathrm{b}}
Time to cross
t=wvbt = \frac{w}{v_\mathrm{b}}
Symbols and units
  • vbv_\mathrm{b}boat velocity through the waterm s−1\mathrm{m\,s^{-1}}
  • vcv_\mathrm{c}velocity of the currentm s−1\mathrm{m\,s^{-1}}
  • wwwidth of the riverm\mathrm{m}

03Sample problems with solutions

Sample 1 · Crossing a river
A boat heads straight across a river 60 m wide at 3.0 m s⁻¹ through the water. The current flows at 1.5 m s⁻¹. Calculate the boat’s resultant velocity, the time to cross and how far downstream it lands.

  1. Resultant velocity:

    v=3.02+1.52=3.35 m s−1,θ=tan⁡−11.53.0=27∘v = \sqrt{3.0^{2} + 1.5^{2}} = 3.35\ \mathrm{m\,s^{-1}}, \quad \theta = \tan^{-1}\frac{1.5}{3.0} = 27^\circ

  2. Only the across-velocity crosses the river; meanwhile the current carries it downstream:

    t=603.0=20 s,x=1.5×20=30 mt = \frac{60}{3.0} = 20\ \mathrm{s}, \qquad x = 1.5 \times 20 = 30\ \mathrm{m}
Answer: 3.4 m s⁻¹ at 27° downstream of straight across; 20 s; 30 m downstream
Unit check: mm s−1=s ✓\frac{\mathrm{m}}{\mathrm{m\,s^{-1}}} = \mathrm{s}\ \checkmark

04Practice problems

  1. P1
    A plane flies north at 200 m s⁻¹ in a wind blowing east at 50 m s⁻¹. Find its resultant velocity.
    Final answer206 m s⁻¹ at 14° east of north
    Answer: 206 m s⁻¹ at 14° east of north
  2. P2
    To land directly opposite, at what angle upstream must the sample’s boat head? What is its speed across?
    Final answer30° upstream; 2.6 m s⁻¹
    Answer: 30° upstream; 2.6 m s⁻¹
  3. P3
    A passenger walks forwards at 1.5 m s⁻¹ in a train moving at 20 m s⁻¹. Find their speed over the ground.
    Final answer21.5 m s⁻¹
    Answer: 21.5 m s⁻¹
  4. P4
    A swimmer crosses a 48 m river at 1.2 m s⁻¹ across a 0.50 m s⁻¹ current. Find the crossing time.
    Final answer40 s
    Answer: 40 s

05Other useful information

Common mistakes

  • ✗ Wrong: Use the resultant speed to find the crossing time.

    ✓ Right: Use only the velocity component across the river.

Exam tips

  • Perpendicular velocities add by Pythagoras, not arithmetic. Say which angle you give.

What the syllabus asks

  • 9702 topic 1: add coplanar vectors (velocities). Coursebook 1.6.

Explore: Practice questions for this lesson