01 · Acceleration vector
A 3 kg particle has resultant 12i − 9j N. Find acceleration vector and magnitude.
Hint
Divide each component by 3.
Worked solution
a = 4i − 3j m/s², with magnitude 5 m/s².
Understand · explore · practise
Vector forces, acceleration and motion
Use vector F = ma with velocity and displacement equations, find bearings and magnitudes, and restart a motion stage when a force is removed.
Before you startForce vector resultants, Newton’s second law and constant acceleration.
01 / Apply the same mass to both force components
Vector equations are two component equations written together.
ΣF = ma, so a = ΣF/m
First add forces on the particle, then divide every component by the positive mass. For constant a, use v = u + at and displacement s = ut + ½at² component by component. Magnitudes are separate calculations.
Constant force from restExplore
A 2 kg particle starts at the origin from rest under constant resultant 6i + 8j N. Take i east and j north. Select time to compare acceleration, velocity and displacement. In this special from-rest case all three directions agree for positive time.
02 / Find the vector acceleration and its magnitude
Do not divide only one component by mass.
A 2 kg particle has forces 9i − 3j N and −3i + 11j N.Worked example
ΣF = 6i + 8j N
Add signed components.
a = 3i + 4j m/s²
Divide both components by 2.
|a| = √(3² + 4²) = 5 m/s²
This is a magnitude, not another vector component.
Bearing of a = tan⁻¹(3/4) = 036.9°
With i east and j north; both components positive.
Watch: force and acceleration share a direction
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Acceleration bearing
For question 1, find the bearing of acceleration to 1 decimal place.
Hint
It points southeast; measure clockwise from north.
Worked solution
Bearing = 180° − tan⁻¹(4/3) = 126.9°. This is not a claim about its current velocity.
03 / Starting from rest gives a special straight-line motion
For t > 0, velocity and displacement are positive multiples of a.
03 · Velocity from rest
The model particle has a = 3i + 4j m/s² and starts from rest. Find velocity and speed at 2 s.
Hint
Use v = at.
Worked solution
v = 6i + 8j m/s; speed = 10 m/s.
04 · Displacement from rest
Find its displacement and distance travelled by 2 s.
Hint
Use s = ½at². The path is straight without reversal.
Worked solution
s = 6i + 8j m. Its magnitude is 10 m, which equals distance here because the from-rest constant-force path is straight.
05 · A specified speed
When does its speed first reach 15 m/s?
Hint
From rest, speed = |a|t for t ≥ 0.
Worked solution
5t = 15 gives t = 3 s.
06 · At launch
What bearing does the model’s velocity have at t = 0?
Hint
The velocity vector is zero.
Worked solution
Undefined. The acceleration has a bearing, but zero velocity has no direction.
04 / A nonzero initial velocity may change the direction relationship
Force direction and velocity direction need not agree.
Now take u = 2i − j m/s and a = 3i + 4j m/s².Worked example
At 3 s: v = (2 + 9)i + (−1 + 12)j = 11i + 11j m/s
Add u rather than assuming it is zero.
Speed = 11√2 m/s; velocity bearing = 045°
Its direction differs from the acceleration bearing 036.9°.
Displacement = 3(2i − j) + 4.5(3i + 4j) = 19.5i + 15j m
The magnitude of this vector is not automatically the distance along a curved path.
07 · Later velocity
With the same u and a, find v at 1 s.
Hint
Add the initial velocity componentwise.
Worked solution
v = 5i + 3j m/s; speed = √34 m/s.
08 · A component event
When is this velocity purely eastward?
Hint
Set the northward component to zero and check the other component.
Worked solution
−1 + 4t = 0 gives t = 0.25 s. The eastward component is 2.75 m/s, so velocity is eastward, not zero.
05 / Infer force or mass from a velocity change
Use acceleration, not velocity, in F = ma.
09 · Unknown force
A 3 kg particle changes velocity from 2i + j to 8i − 3j m/s over 2 s under constant resultant. Find that resultant.
Hint
First divide the velocity change by time.
Worked solution
a = (6i − 4j)/2 = 3i − 2j m/s². Resultant = 9i − 6j N.
10 · Unknown mass
Resultant force is 12i + 16j N and acceleration is 3i + 4j m/s². Find mass and verify both components.
Hint
The same mass must work in both equations.
Worked solution
12 = 3m gives m = 4 kg; 16 = 4m confirms the same value.
11 · Inconsistent data
Could resultant 12i + 16j N produce acceleration 3i + 5j m/s² for any positive mass?
Hint
Compare the mass implied by each component.
Worked solution
No. The i equation requires 4 kg while the j equation requires 3.2 kg. The force and acceleration vectors are not parallel.
06 / A removed force changes acceleration but not velocity instantaneously
Carry the current state into a new local clock.
Recalculate the resultant after the change.
Removing a finite force does not create an instantaneous velocity jump in this model. The new force changes the rate at which velocity changes. Use the old final velocity as the new initial velocity.
12 · New acceleration
A 2 kg particle starts from rest with forces 6i N and 8j N. After 2 s the 6i force is removed. Find the new acceleration and velocity at the instant of removal.
Hint
Use the old acceleration to reach the event, then the remaining force for the new acceleration.
Worked solution
At removal v = 6i + 8j m/s. New acceleration = 4j m/s². The eastward velocity does not suddenly disappear.
13 · One second later
Find velocity 1 s after the change in question 12.
Hint
Use local time τ = 1 with the new acceleration.
Worked solution
v = (6i + 8j) + 4j = 6i + 12j m/s.
14 · Total displacement
Find displacement from the original launch point 1 s after the change.
Hint
Add displacement in the first stage to displacement in the second.
Worked solution
First stage: 6i + 8j m. Second stage: (6i + 8j) × 1 + ½(4j) × 1² = 6i + 10j m. Total = 12i + 18j m.
07 / Distinguish position, displacement and distance
The origin and path matter.
Displacement is the change in position. If the particle starts at position r₀, its later position is r₀ + ut + ½at². A displacement magnitude is the straight-line separation of endpoints; a curved or reversing path can have a greater distance travelled. Do not substitute vector magnitudes into scalar SUVAT unless a justified one-dimensional model applies.
For constant force from rest, the path follows the fixed acceleration direction, so the magnitude shortcut is valid over positive time. With a nonparallel initial velocity, use components and interpret the result carefully.
08 / Keep components until the final magnitude or bearing
Restart the model whenever its force set changes.
Add forces, divide by mass, and evolve each velocity and displacement component. Distinguish acceleration bearing from velocity bearing. At a force change, preserve the state and calculate the new acceleration.
Section 1 of 8 · Apply the same mass to both force components