SPEC 7.3 • KINEMATICS
SUVAT Equations in 2D Using Vectors
Extending constant acceleration formulae to two dimensions with vector notation
Vector SUVAT Equations
For motion in two dimensions with constant acceleration, we extend the SUVAT equations using vectors:
Key Vector Equations
What These Mean
v = velocity vector at time t
u = initial velocity vector
a = constant acceleration vector
r = position/displacement vector at time t
t = time (a scalar, same for both x and y directions)
Vector Notation
Vectors can be written in two equivalent forms:
i-j Notation
Example:
m s⁻¹
Column Vector
Example:
m s⁻¹
Standard Convention
i = unit vector in positive x-direction
j = unit vector in positive y-direction
These are perpendicular directions in 2D space
Finding Velocity at Time t
Use v = u + at to find the velocity vector at any time:
Method
Simply add the vectors:
Key Point
The vector equation handles both the x and y components simultaneously. Each component follows its own 1D SUVAT equation with the same time t.
Worked Example01
A particle has initial velocity m s⁻¹ and constant acceleration m s⁻². Find the velocity when t = 1.5 s.
1
Use v = u + at
2
3
Answer m s⁻¹ or (8i − 5j) m s⁻¹
Finding Position at Time t
Use r = ut + ½at² to find the position vector:
Method
Interpretation
The position vector r gives the displacement from the starting point. The x-component tells you how far in the x-direction, and the y-component tells you how far in the y-direction.
Worked Example02
A particle starts from the origin with velocity m s⁻¹ and acceleration m s⁻². Find the position vector after 3 seconds.
1
Use r = ut + ½at² with t = 3
2
3
Answer m or (27i − 21j) m
Resultant Force and Acceleration
In 2D motion, the acceleration vector comes from the resultant force acting on the particle:
Newton's Second Law in 2D
If multiple forces act, find the resultant first:
Finding Acceleration from Force
Given the resultant force and mass:
Worked Example03
A particle of mass 2 kg is acted upon by forces N and N. Find the acceleration of the particle.
1
Find resultant force: N
2
Use F = ma, so a = F/m
3
m s⁻²
Answer m s⁻² or (2i + 6j) m s⁻²
Speed and Distance
Speed and distance are scalars found from the magnitude of vectors:
Magnitude Formulas
Speed (magnitude of velocity):Distance (magnitude of displacement):
Direction of Motion
The angle θ to the positive x-direction can be found using:
Worked Example04
A particle has initial velocity (15 i + 20j) m s⁻¹ and acceleration −10j m s⁻². Find the speed of the particle after 3 seconds.
1
Find velocity: v = u + at = (15i + 20j) + (−10j)(3)
2
v = 15i + 20j − 30j = (15i − 10j) m s⁻¹
3
Speed = |v| = √(15² + (−10)²) = √(225 + 100) = √325
4
Speed = √325 = 5√13 m s⁻¹
AnswerSpeed = 5√13 m s⁻¹ (≈ 18.0 m s⁻¹)
Special Conditions
Exam questions often ask when a particle satisfies certain conditions:
Common Conditions
ConditionMathematical Meaning"Moving parallel to i"y-component of velocity = 0, i.e., vᵧ = 0"Moving parallel to j"x-component of velocity = 0, i.e., vₓ = 0"Moving parallel to "Velocity is a scalar multiple: vᵧ/vₓ = b/a
Method
Set up the condition as an equation and solve for t. Then substitute back to find other quantities if needed.
Worked Example05
A particle is projected from O with velocity (5 i + 14j) m s⁻¹ and acceleration −10j m s⁻². Find the time when the particle is moving parallel to i.
1
"Parallel to i" means the y-component of velocity = 0
2
Find velocity: v = (5i + 14j) + (−10j)t = 5i + (14 − 10t)j
3
Set y-component = 0: 14 − 10t = 0
4
Solve: 10t = 14, so t = 1.4 s
Answert = 1.4 seconds
Worked Example06
A particle is projected with velocity m s⁻¹ and acceleration m s⁻². Find the time when the particle is moving parallel to the vector .
1
Velocity at time t:
2
Parallel to means vᵧ/vₓ = −1/3
3
So: (24 − 10t)/12 = −1/3
4
Cross multiply: 3(24 − 10t) = −12
72 − 30t = −12
30t = 84
t = 2.8 s
Answert = 2.8 seconds
Verifying a Point is Reached
To check if a particle passes through a specific point:
Method
Use one component (usually x) to find the time t
Substitute this t into the other component (y)
Check if the calculated y matches the given y-coordinate
If They Match
The particle passes through the point at the calculated time.
If They Don't Match
The particle does NOT pass through that point. State clearly that the coordinates are inconsistent.
Worked Example07
A particle P is projected from O with velocity m s⁻¹ and acceleration m s⁻². Find the time when P passes through the point with position vector m.
1
Position:
2
From x-component: 20t = 40 → t = 2 s
3
Check y-component at t = 2: y = 30(2) − 5(4) = 60 − 20 = 40 m
4
But we need y = 35 m, and we got y = 40 m. These don't match!
AnswerThe particle does not pass through . At t = 2 s when x = 40 m, the y-coordinate is 40 m, not 35 m.
Practice
Question 1: A particle has initial velocity m s⁻¹ and acceleration m s⁻². Find the velocity when t = 2 s.
Solution:
m s⁻¹
Practice
Question 2: A particle starts from O with velocity (8i + 6j) m s⁻¹ and acceleration (2i − 10j) m s⁻². Find:
(a) the position after 2 seconds
(b) the distance from O after 2 seconds
Solution:
(a) r = (8i + 6j)(2) + ½(2i − 10j)(4)
r = (16i + 12j) + (4i − 20j) = (20i − 8j) m
(b) Distance = |r| = √(20² + 8²) = √(400 + 64) = √464 = 4√29 m ≈ 21.5 m
Practice
Question 3: A particle is projected with velocity m s⁻¹ and acceleration m s⁻². Find the time when the particle is moving parallel to the vector .
Solution:
Velocity:
Parallel to means vᵧ/vₓ = −1/1 = −1
So: (18 − 10t)/6 = −1
18 − 10t = −6
10t = 24
t = 2.4 seconds
Summary
Vector SUVAT Equations
EquationUse For
v = u + a tFinding velocity at time t
r = u t + ½ a t²Finding position at time t
F = m aFinding acceleration from resultant force
Key Points
Use i-j notation or column vectors
Acceleration comes from resultant force: a = F/m
"Parallel to i" means y-component of velocity = 0
Speed = |v| = √(vₓ² + vᵧ²)
To verify a point, check both components give consistent time
Next: Vector SUVAT Equations