01 · Speed
Find the speed for v=−8i+6j m/s.
Hint
Square both components.
Worked solution
√(64+36)=10 m/s.
Understand · explore · practise
Vector magnitudes and directions in mechanics
Find speed and other vector magnitudes from perpendicular components, choose the correct quadrant and state the angle reference explicitly.
Before you startPythagoras, right-triangle trigonometry and scalars versus vectors.
01 / Components describe one vector, not two separate journeys
Use perpendicular unit directions consistently.
For v=ai+bj, its magnitude is √(a²+b²).
The units remain those of v. For velocity the magnitude is speed; for acceleration it is acceleration magnitude. A magnitude alone does not give direction.
Same magnitude, different quadrantExplore
i points right and j points up. The selected vector is a velocity in m/s. Direction angles are anticlockwise from positive i, in degrees from 0 to less than 360.
02 / Use Pythagoras on perpendicular components
Negative signs disappear when squared, but still affect direction.
A velocity is 3i+4j m/s.Worked example
|v|=√(3²+4²)=5 m/s
This is the speed.
For −3i+4j, |v| is still 5 m/s
The direction changes to the upper-left quadrant.
Do not add 3+4 to obtain speed
The components are perpendicular, not collinear.
Watch: components form a right triangle
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Acceleration
Find the magnitude of a=6i−8j m/s².
Hint
Keep acceleration units.
Worked solution
√(36+64)=10 m/s². This is not a speed.
03 · Force
Find the magnitude of F=−5i+12j N.
Hint
Use the same geometric rule with force units.
Worked solution
√(25+144)=13 N.
03 / Name the axis from which you measure an angle
An acute triangle angle is not automatically the requested direction.
For v=3i+4j m/s:Worked example
Angle above positive i: θ=tan⁻¹(4/3)≈53.13°
Opposite component is 4 and adjacent component is 3.
Angle to the right of positive j: φ=tan⁻¹(3/4)≈36.87°
The reference axis has changed.
θ+φ=90°
These are complementary angles in this first-quadrant triangle.
04 · Axis wording
Why should an answer of 36.87° alone be avoided here?
Hint
A direction requires a reference and a sense.
Worked solution
It must specify the reference axis and side, such as 36.87° to the right of positive j. Otherwise it can be confused with the angle above positive i.
05 · Units
Does the direction angle carry the velocity unit m/s?
Hint
Angles and vector magnitudes are different quantities.
Worked solution
No. Report the angle in degrees or radians as appropriate, and the speed separately in m/s.
04 / Use signs to place the vector before finding the angle
Inverse tangent of a ratio may omit quadrant information.
06 · Upper left
Find the anticlockwise direction from positive i for −3i+4j.
Hint
Use the acute reference angle 53.13° in quadrant II.
Worked solution
180°−53.13°≈126.87°.
07 · Lower left
Find that direction for −3i−4j.
Hint
Both components are negative.
Worked solution
180°+53.13°≈233.13°.
08 · Lower right
Find that direction for 3i−4j, using 0≤θ<360°.
Hint
Use quadrant IV.
Worked solution
360°−53.13°≈306.87°. A signed angle of −53.13° describes the same direction but is outside the requested interval.
05 / Handle zero components directly
Avoid division by zero in a tangent ratio.
09 · Vertical
Find the speed and anticlockwise direction for v=5j m/s.
Hint
The vector is on positive j.
Worked solution
Speed=5 m/s and direction=90° from positive i.
10 · Horizontal
Find the magnitude and direction for a=−7i m/s².
Hint
The vector lies on negative i.
Worked solution
Magnitude=7 m/s² and direction=180° from positive i.
11 · Zero
What is the direction angle of a zero velocity vector?
Hint
There is no distinguished direction.
Worked solution
It has zero speed but no unique direction angle. Do not assign an angle using a zero-over-zero ratio.
06 / Subtract positions before finding displacement magnitude
Retain the units of the original coordinates.
12 · Displacement
A point moves from position (2,−1) m to (−4,7) m. Find displacement and its magnitude.
Hint
Final position minus initial position.
Worked solution
Displacement=(−6,8) m = −6i+8j m. Its magnitude is √(36+64)=10 m.
13 · Direction from north
If i is east and j north, describe that displacement’s direction relative to north.
Hint
It points west of north; use tan⁻¹(6/8).
Worked solution
Approximately 36.87° west of north. The negative east component determines the westward side.
07 / Distinguish magnitude, component and total path length
They answer different questions.
14 · Component versus speed
A velocity has i-component −8 m/s. Can its speed be less than 8 m/s?
Hint
The other squared component is nonnegative.
Worked solution
No. Its speed is at least 8 m/s, with equality if all perpendicular components are zero.
15 · Scaling
If a velocity vector doubles, what happens to its speed?
Hint
Apply the magnitude formula or homogeneity.
Worked solution
Its speed doubles: |2v|=2|v|. Multiplying by −2 would also double the speed but reverse the direction.
16 · Distance claim
Does the 10 m displacement magnitude in question 12 prove the object travelled exactly 10 m?
Hint
The intermediate path is not specified.
Worked solution
No. It is the straight-line separation of the endpoints. The actual distance travelled can be larger.
08 / Sketch the quadrant, then calculate
State units and angle convention.
Use Pythagoras for magnitudes, keeping the physical unit. For direction, identify the quadrant and the reference axis before using trigonometry. Handle vectors on an axis directly and leave the zero vector without a unique angle. Displacement magnitude is not generally total path length.
Section 1 of 8 · Components describe one vector, not two separate journeys