SME
Dynamics · Problem type
Relative velocity of two moving particles
Use this for “Car A drives north at 20 m/s and car B west at 15 m/s. Find B’s velocity relative to A.”
Diagram
vB: 15 m/s at 180°, so vB = ⟨−15, 0⟩.
−vA: 20 m/s at 270°, so −vA = ⟨0, −20⟩.
Tip to tail: vB/A = ⟨−15, −20⟩
|vB/A| = 25 m/s at 233.1°.
This is an example. Type your own values to start a new problem.
vₓ (m/s)
=
vB (m/s)
cos
θB
°
−
vA (m/s)
cos
θA
°
vᵧ (m/s)
=
vB (m/s)
sin
θB
°
−
vA (m/s)
sin
θA
°
θA is car A’s heading; the picture draws −vA, so that arrow points the opposite way, at θA + 180°.
vB/A
Relative speed
worked out (not typed)
θB/A
Direction of vB/A
worked out (not typed) · °
Formulas
vₓ = vB × cos(θB) − vA × cos(θA)
−15 = 15 × cos(180°) − 20 × cos(90°)
vᵧ = vB × sin(θB) − vA × sin(θA)
−20 = 15 × sin(180°) − 20 × sin(90°)
vB/A
=
√
vₓ
2
+
vᵧ
2
25
=
√(
(−15)
2
+
(−20)
2
)
θB/A = atan2(vᵧ, vₓ)
233.1301 = atan2(−20, −15)
Assumptions
•
Directions are measured counterclockwise from east (0°), so north is 90° and west is 180°.
•
vB/A = vB − vA: what someone riding with A sees B doing. Subtracting vA adds −vA tip to tail.
•
Both velocities are measured from the same fixed ground, and A does not rotate.
Step-by-step
WE KNOW
vA = 20 m/s, θA = 90°, vB = 15 m/s, θB = 180°
FIND
east part of vB/A (vₓ), north part of vB/A (vᵧ), relative speed (vB/A), direction of vB/A (θB/A)
Step 1 · Find east part of vB/A (vₓ)
Use vₓ = vB × cos(θB) − vA × cos(θA)
Subtract the velocities part by part: B’s east part less A’s east part.
vₓ = 15 × cos(180°) − 20 × cos(90°)
vₓ = 15 × (−1) − 20 × 0
vₓ = −15 m/s
Step 2 · Find north part of vB/A (vᵧ)
Use vᵧ = vB × sin(θB) − vA × sin(θA)
The same for the north parts: B’s less A’s.
vᵧ = 15 × sin(180°) − 20 × sin(90°)
vᵧ = 15 × 0 − 20
vᵧ = 0 − 20
vᵧ = −20 m/s
Step 3 · Find relative speed (vB/A)
Use vB/A = √(vₓ² + vᵧ²)
The two parts are at right angles, so the length is their Pythagorean sum.
vB/A
=
√(
(−15)
2
+
(−20)
2
)
vB/A
=
√
225 + 400
vB/A
=
√
625
vB/A = 25 m/s
Step 4 · Find direction of vB/A (θB/A)
Use θB/A = atan2(vᵧ, vₓ)
The angle from east: tan⁻¹ of north over east, turned into the quadrant the parts point to.
θB/A = atan2(vᵧ, vₓ) + 360°
θB/A = atan2(−20, −15) + 360°
θB/A = 233.1301°
Check
Put the numbers back in. Both sides should match.
−15 = 15 × cos(180°) − 20 × cos(90°) ✓
−20 = 15 × sin(180°) − 20 × sin(90°) ✓
25
=
√(
(−15)
2
+
(−20)
2
)
✓
233.13 = atan2(−20, −15) + 360° ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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