01 · A shared time
From the origin, u = (1, −1, 2) and a = (2, 0, −1), in SI units. Find r and v at t = 3.
Hint
Use v = u + 3a and r = 3u + 4.5a.
Worked solution
v = (7,−1,−1) m s⁻¹; r = (12,−3,1.5) m.
Understand · explore · practise
Motion with 3D vectors
Use constant vector acceleration to find velocity and position, distinguish distance from displacement and use vertical velocity to decide ascent or descent.
Before you startVector arithmetic, magnitude, differentiation and constant-acceleration equations.
01 / Apply the same time to all three components
Constant acceleration gives vector equations for velocity and position.
v = u + at; r = r₀ + ut + ½at²
u is initial velocity; r₀ is initial position; t is elapsed time.
These equations require constant vector acceleration. Position, velocity and acceleration are different quantities with different units. The three component motions share one clock.
Inspect position and velocity at your chosen timeExplore
Start at the origin with constant a = (1, 0, 2) m s⁻². Positive z is upwards.
r = (0.5, 0, 1) m; |r| = 1.118034 m.
v = (1, 0, 2) m s⁻¹; speed = 2.236068 m s⁻¹.
Ascending: vertical velocity is positive.
From rest, motion follows the fixed direction a without reversal: distance = |r|.
The trajectory is an oblique projection. The amber velocity indicator uses a separate visual scale; its direction, not its drawn length relative to displacement, is meaningful. No automatic time advance.
02 / Calculate x, y and z separately
Keep the vectors until the calculation is complete.
r₀ = (1, −2, 3) m, u = (2, 1, −1) m s⁻¹, a = (1, −2, 2) m s⁻². Find r and v after 2 s.Worked example
v = (2,1,−1) + 2(1,−2,2) = (4,−3,3) m s⁻¹
Use v = u + at.
r = (1,−2,3) + 2(2,1,−1) + 2(1,−2,2)
Here ½t² = 2.
r = (7,−4,5) m
Add each component.
02 · Recover acceleration
Velocity changes from (1, 2, 3) to (5, −2, 7) m s⁻¹ in 2 s under constant acceleration. Find a.
Hint
Use (v − u)/t.
Worked solution
a = (2,−2,2) m s⁻².
03 / Speed is the magnitude of velocity
Do not add the velocity components to obtain speed.
v = (4, −3, 3) m s⁻¹.Worked example
Speed = √(4² + (−3)² + 3²) = √34 m s⁻¹
Speed is nonnegative.
The negative y component describes direction
It does not make the speed negative.
03 · Speed
Find the speed when v = (2, −3, 6) m s⁻¹.
Hint
Take the magnitude.
Worked solution
√(4 + 9 + 36) = 7 m s⁻¹.
04 · Instantaneous rest
Is a particle at rest when v = (0, 3, 0) m s⁻¹?
Hint
All components must vanish.
Worked solution
No. Its speed is 3 m s⁻¹.
04 / From rest, constant acceleration fixes the direction of motion
For t ≥ 0, displacement is a nonnegative multiple of a.
u = 0 ⇒ s = ½at² and v = at
Here s = r − r₀ is displacement.
a = (2, −1, 2) m s⁻², starting from rest. Find the distance travelled after 4 s.Worked example
|a| = 3 m s⁻²
Find the constant acceleration magnitude.
s = 8a = (16,−8,16) m
For t = 4, ½t² = 8.
Distance = |s| = 8 × 3 = 24 m
The path is straight and there is no reversal after starting from rest.
05 · Rest case
Starting from rest with a = (1,2,2) m s⁻², find speed and distance after 3 s.
Hint
|a| = 3; use |v| = t|a| and distance = ½t²|a|.
Worked solution
Speed = 9 m s⁻¹; distance = 13.5 m.
05 / Nonparallel initial velocity and acceleration can give a curved path
Constant acceleration does not by itself imply straight motion.
u = (0, 2, 0), a = (1, 0, 2), starting at the origin.Worked example
r(t) = (t²/2, 2t, t²)
The y component changes linearly; x and z change quadratically.
At t = 1, r = (0.5,2,1); at t = 2, r = (2,4,4)
These displacements are not scalar multiples.
The motion is not along one fixed line
The path curves even though a is constant.
06 · Check straightness
In this example, is r(2) equal to 2r(1)? What does that comparison alone show?
Hint
Compute 2r(1) = (1,4,2).
Worked solution
No. This alone rules out doubling at those times. To show the two positions are not collinear with the origin, note their component ratios are 4, 2, 4, so no single scalar relates them.
06 / Distance accumulates path length; displacement joins endpoints
Even straight-line motion can reverse.
u = (0,0,4) m s⁻¹ and a = (0,0,−2) m s⁻². Start at the origin and consider 0 ≤ t ≤ 4.Worked example
z(t) = 4t − t²; vz = 4 − 2t
The motion is vertical.
At t = 2, vz = 0 and z = 4 m
The particle reaches its highest point.
At t = 4, z = 0
Net displacement is zero.
Distance travelled = 4 + 4 = 8 m
It travelled up and back down.
Watch: returning to the start does not erase distance
Pause, replay or seek freely. The notes explain the same idea and stay in view.
07 · Turning motion
For this motion, find distance and displacement over 0 ≤ t ≤ 3.
Hint
It rises to z = 4, then ends at z = 3.
Worked solution
Distance = 4 + 1 = 5 m; displacement = (0,0,3) m.
08 · General distance expression
How is total distance over [0,T] obtained from a velocity vector v(t)?
Hint
Integrate the speed.
Worked solution
Distance = ∫₀ᵀ |v(t)| dt. In general this is not equal to |r(T) − r(0)|.
07 / Use vertical velocity to decide ascent or descent
With z upwards, vz > 0 means ascending and vz < 0 descending.
u = (0,0,−4) and a = (1,0,2), in SI units.Worked example
vz = −4 + 2t
The vertical acceleration is positive throughout.
For 0 ≤ t < 2, vz < 0
The particle is descending.
At t = 2, vz = 0, but vx = 2
It is momentarily moving horizontally, not at rest.
For t > 2, vz > 0
It is ascending.
09 · Ascent threshold
If initial vertical velocity is −6 m s⁻¹ and constant vertical acceleration is 3 m s⁻², when does ascent begin?
Hint
Solve −6 + 3t = 0.
Worked solution
The vertical velocity is zero at 2 s and positive for t > 2 s. Ascent begins after that instant.
10 · Height and motion
Can a particle with z > 0 still be descending?
Hint
Position and velocity describe different things.
Worked solution
Yes. It is above the chosen origin, but descends if vz < 0.
08 / Solve a requested condition and check every component
A zero component is not necessarily a zero vector.
u = (−2, 1, −4), a = (1, 0, 2). Does the particle ever come to rest?Worked example
vx = −2 + t and vz = −4 + 2t vanish at t = 2
Two components agree on a candidate.
vy = 1 at every time
The third component never vanishes.
The particle is never at rest
At t = 2 it has velocity (0,1,0).
11 · Reaching a point
r(t) = (t, 2t, t²) m for t ≥ 0. Does the particle reach (2,4,5)?
Hint
The first two coordinates suggest t = 2. Check the third.
Worked solution
No. At t = 2 the position is (2,4,4).
12 · Reaching a plane
For r(t) = (t,2t,t²), when does the particle reach z = 9? State its position.
Hint
Use t ≥ 0.
Worked solution
t = 3 s; r = (3,6,9) m.
09 / Use scalar equations along a fixed coordinate axis
Do not substitute arbitrary vector magnitudes into signed one-dimensional formulas.
vx² = ux² + 2ax sx
This holds for the x components when ax is constant; analogous equations hold for y and z.
A scalar speed-and-distance formula may be used for genuinely straight motion with consistent signed quantities, accounting for reversals. For nonparallel u and a, use the vector equations first. There is no general rule |v|² = |u|² + 2|a||s|.
13 · A false magnitude rule
Take u = (0,1,0), a = (1,0,0), t = 1. Compare |v|² with |u|² + 2|a||s|.
Hint
v = (1,1,0), s = (1/2,1,0).
Worked solution
|v|² = 2, whereas the proposed right side is 1 + √5. They differ.
10 / Convert a constant resultant to acceleration before using motion equations
A constant force and constant mass give constant vector acceleration.
A 2 kg particle starts from rest under constant resultant (4, −2, 4) N.Worked example
a = (2, −1, 2) m s⁻²
Use R/m.
After 3 s: v = (6,−3,6) m s⁻¹
Multiply a by time.
s = (9,−4.5,9) m
Use ½at².
Distance = 13.5 m
This is the straight, no-reversal rest case.
14 · From force to position
A 4 kg particle at the origin starts from rest under R = (0,8,−4) N. Find its position at t = 2 s.
Hint
a = (0,2,−1); ½t² = 2.
Worked solution
r = (0,4,−2) m.
15 · Check constancy
If the resultant force changes with time, may you automatically use r = r₀ + ut + ½at² with one fixed a?
Hint
The derivation assumes constant acceleration.
Worked solution
No. Find a(t) from the force and integrate as needed, or justify a suitable constant approximation.
11 / Keep position, displacement, velocity and distance distinct
State the constant-acceleration assumption before calculating.
- Use a shared elapsed time in every component.
- Take |v| for speed and |s| for displacement magnitude.
- Distance equals |s| for a straight path without reversal.
- From rest under constant a, that condition holds for t ≥ 0.
- Use the sign of vertical velocity for ascent or descent.
- Check all components when solving a vector condition.
16 · Final interpretation
At one instant v = (3,0,0) m s⁻¹ and a = (0,0,2) m s⁻². Is the particle ascending at that instant?
Hint
Inspect vz, not az.
Worked solution
No: vz = 0, so it is moving horizontally at that instant. If this acceleration continues, it will acquire upward velocity immediately afterwards.
Section 1 of 11 · Apply the same time to all three components