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Displacement-Velocity-Acceleration

CSEC Additional Mathematics Essential Knowledge: The relationships between displacement, velocity, and acceleration form the foundation of kinematics. Understanding how these quantities are connected through differentiation and integration is crucial for analyzing motion along a straight line. This topic combines calculus concepts with practical physics applications.

Key Concept: Displacement, velocity, and acceleration are mathematically related: Velocity is the derivative of displacement with respect to time, and acceleration is the derivative of velocity with respect to time. Conversely, displacement is the integral of velocity, and velocity is the integral of acceleration.

Part 1: Fundamental Relationships

📊

The Calculus Connection

🧮
Differentiation Relationships
\[v = \frac{ds}{dt} \quad \text{and} \quad a = \frac{dv}{dt} = \frac{d^2s}{dt^2}\]
D
Differentiate Displacement → Velocity
Slope of displacement-time graph = Velocity
D
Differentiate Velocity → Acceleration
Slope of velocity-time graph = Acceleration
📈
Integration Relationships
\[s = \int v \, dt \quad \text{and} \quad v = \int a \, dt\]
∫
Integrate Acceleration → Velocity
Area under acceleration-time graph = Change in velocity
∫
Integrate Velocity → Displacement
Area under velocity-time graph = Displacement
📝 Example 1: From Displacement to Acceleration

The displacement of a particle is given by \(s = 2t^3 – 5t^2 + 3t + 1\) meters. Find:

(a) Velocity as a function of time

(b) Acceleration as a function of time

(c) Velocity and acceleration at \(t = 2\) seconds

1
Part (a): Differentiate s to get v: \(v = \frac{ds}{dt} = 6t^2 – 10t + 3\) m/s
2
Part (b): Differentiate v to get a: \(a = \frac{dv}{dt} = 12t – 10\) m/s²
3
Part (c): At t = 2: \(v = 6(2)^2 – 10(2) + 3 = 24 – 20 + 3 = 7\) m/s
4
\(a = 12(2) – 10 = 24 – 10 = 14\) m/s²

Part 2: Graphical Relationships

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Graphs of Motion and Their Connections

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Displacement-Time Graph

The slope (gradient) at any point on a displacement-time graph gives the instantaneous velocity.

Displacement-Time Graph: \(s = t^3 – 6t^2 + 9t\)

Key Points:

  • When slope = 0 (horizontal tangent), velocity = 0 (particle momentarily at rest)
  • Positive slope → positive velocity
  • Negative slope → negative velocity
  • Curvature indicates acceleration
🚀
Velocity-Time Graph

The slope gives acceleration, and the area under the curve gives displacement.

Velocity-Time Graph: \(v = 3t^2 – 12t + 9\)

Key Points:

  • Slope = acceleration
  • Area above time axis = positive displacement
  • Area below time axis = negative displacement
  • When v = 0, particle changes direction
⚡
Acceleration-Time Graph

The area under the curve gives change in velocity.

Acceleration-Time Graph: \(a = 6t – 12\)

Key Points:

  • Area above time axis → increase in velocity
  • Area below time axis → decrease in velocity
  • When a = 0, velocity is at a maximum or minimum
  • Constant acceleration = horizontal line

All Three Graphs Together

Displacement-Time
Velocity-Time
Acceleration-Time

Note: These graphs show the relationships for \(s = t^3 – 6t^2 + 9t\). Notice how:

  • The velocity graph is the derivative of the displacement graph
  • The acceleration graph is the derivative of the velocity graph
  • When displacement has maximum/minimum points, velocity = 0
  • When velocity has maximum/minimum points, acceleration = 0

Part 3: Practical Applications

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Solving Real-World Problems

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Common Problem Types
Given Find Method
Displacement function s(t) Velocity and acceleration Differentiate
Velocity function v(t) Displacement and acceleration Integrate for s, differentiate for a
Acceleration function a(t) Velocity and displacement Integrate (need initial conditions)
Graph of one quantity Graphs of related quantities Use slope/area relationships
📝 Example 2: From Acceleration to Displacement

A particle moves along a straight line with acceleration \(a = 12t – 6\) m/s². Given that when \(t = 0\), \(v = 2\) m/s and \(s = 1\) m, find:

(a) Velocity as a function of time

(b) Displacement as a function of time

(c) When the particle is at rest

1
Part (a): Integrate acceleration: \(v = \int (12t – 6) dt = 6t^2 – 6t + C\)
2
Use initial condition: When \(t = 0\), \(v = 2\): \(2 = 0 – 0 + C \Rightarrow C = 2\)
3
So \(v = 6t^2 – 6t + 2\) m/s
4
Part (b): Integrate velocity: \(s = \int (6t^2 – 6t + 2) dt = 2t^3 – 3t^2 + 2t + D\)
5
Use initial condition: When \(t = 0\), \(s = 1\): \(1 = 0 – 0 + 0 + D \Rightarrow D = 1\)
6
So \(s = 2t^3 – 3t^2 + 2t + 1\) meters
7
Part (c): At rest when v = 0: \(6t^2 – 6t + 2 = 0\)
8
Solve quadratic: Divide by 2: \(3t^2 – 3t + 1 = 0\)
9
Discriminant: \((-3)^2 – 4(3)(1) = 9 – 12 = -3 < 0\)
10
Conclusion: No real solutions, so particle is never at rest
📝 Example 3: Interpreting Graphs

The velocity-time graph below shows a particle’s motion:

• From t=0 to t=3: v increases from 0 to 6 m/s (straight line)

• From t=3 to t=7: v decreases from 6 to -2 m/s (straight line)

• From t=7 to t=10: v increases from -2 to 4 m/s (straight line)

1
Find acceleration in each segment:
0-3s: \(a = \frac{6-0}{3-0} = 2\) m/s²
3-7s: \(a = \frac{-2-6}{7-3} = \frac{-8}{4} = -2\) m/s²
7-10s: \(a = \frac{4-(-2)}{10-7} = \frac{6}{3} = 2\) m/s²
2
Find displacement (area under v-t graph):
0-3s: Area = triangle = \(\frac{1}{2} \times 3 \times 6 = 9\) m
3-7s: Area = trapezoid = \(\frac{1}{2} \times (6 + (-2)) \times 4 = \frac{1}{2} \times 4 \times 4 = 8\) m
7-10s: Area = trapezoid = \(\frac{1}{2} \times ((-2) + 4) \times 3 = \frac{1}{2} \times 2 \times 3 = 3\) m
3
Total displacement: \(9 + 8 + 3 = 20\) m
4
When does particle change direction? When v changes sign: between t=3 and t=7
Solve for when v=0: From t=3 to t=7, v decreases from 6 to -2 at constant rate
Equation: \(v = 6 – 2(t-3)\)
Set v=0: \(0 = 6 – 2(t-3) \Rightarrow 2(t-3) = 6 \Rightarrow t-3 = 3 \Rightarrow t = 6\) s

Part 4: Special Cases and Important Points

🎯

Key Moments in Motion

Instantaneously at Rest

When: \(v = 0\)

On s-t graph: Horizontal tangent (slope = 0)

On v-t graph: Graph crosses time axis

Interpretation: Particle changes direction or momentarily stops

Maximum/Minimum Displacement

When: \(v = 0\) and \(a \neq 0\)

On s-t graph: Turning point (local max/min)

Test: If \(a < 0\) at this point → maximum displacement

If \(a > 0\) at this point → minimum displacement

Constant Velocity

When: \(a = 0\)

On v-t graph: Horizontal line

On s-t graph: Straight line (constant slope)

Interpretation: Uniform motion

📏
Distance vs Displacement

Important Distinction:

  • Displacement: Net change in position (vector, can be positive or negative)
  • Distance: Total path length traveled (scalar, always positive or zero)

Calculating distance from velocity: \( \text{Distance} = \int |v| \, dt \)

This means you must consider the absolute value of velocity and integrate separately over intervals where v is positive and negative.

📝 Example 4: Distance vs Displacement

A particle moves with velocity \(v = t^2 – 4t + 3\) m/s. Find:

(a) Displacement from t=0 to t=4

(b) Distance traveled from t=0 to t=4

1
Part (a): Displacement = \(\int_0^4 v \, dt\):
\(\int_0^4 (t^2 – 4t + 3) dt = \left[ \frac{t^3}{3} – 2t^2 + 3t \right]_0^4\)
2
\(= \left( \frac{64}{3} – 32 + 12 \right) – 0 = \frac{64}{3} – 20 = \frac{64}{3} – \frac{60}{3} = \frac{4}{3}\) m
3
Part (b): First find when v=0: \(t^2 – 4t + 3 = 0 \Rightarrow (t-1)(t-3) = 0 \Rightarrow t=1, 3\)
4
Test intervals:
On [0,1]: Test t=0.5: v positive
On [1,3]: Test t=2: v negative
On [3,4]: Test t=3.5: v positive
5
Distance = \(\int_0^1 v \, dt – \int_1^3 v \, dt + \int_3^4 v \, dt\)
6
\(\int_0^1 v \, dt = \left[ \frac{t^3}{3} – 2t^2 + 3t \right]_0^1 = \frac{1}{3} – 2 + 3 = \frac{1}{3} + 1 = \frac{4}{3}\)
7
\(\int_1^3 v \, dt = \left[ \frac{t^3}{3} – 2t^2 + 3t \right]_1^3 = (9 – 18 + 9) – (\frac{1}{3} – 2 + 3) = 0 – (\frac{1}{3} + 1) = -\frac{4}{3}\)
8
\(\int_3^4 v \, dt = \left[ \frac{t^3}{3} – 2t^2 + 3t \right]_3^4 = (\frac{64}{3} – 32 + 12) – (9 – 18 + 9) = \frac{4}{3} – 0 = \frac{4}{3}\)
9
Distance: \(\frac{4}{3} – (-\frac{4}{3}) + \frac{4}{3} = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4\) m

Quiz: Test Your Understanding

Displacement-Velocity-Acceleration Quiz
Question 1: A particle’s displacement is given by \(s = 4t^3 – 12t^2 + 9t\). Find when it is momentarily at rest.
Answer:
\(v = \frac{ds}{dt} = 12t^2 – 24t + 9\)
Set \(v = 0\): \(12t^2 – 24t + 9 = 0\)
Divide by 3: \(4t^2 – 8t + 3 = 0\)
Factor: \((2t-1)(2t-3) = 0\)
So \(t = 0.5\) s and \(t = 1.5\) s
Question 2: A particle moves with acceleration \(a = 4 – 2t\). Given that when \(t=0\), \(v=6\) m/s and \(s=0\) m, find the displacement function.
Answer:
First integrate a to get v: \(v = \int (4 – 2t) dt = 4t – t^2 + C\)
Using \(v(0)=6\): \(6 = 0 – 0 + C \Rightarrow C = 6\)
So \(v = 4t – t^2 + 6\)
Now integrate v to get s: \(s = \int (4t – t^2 + 6) dt = 2t^2 – \frac{t^3}{3} + 6t + D\)
Using \(s(0)=0\): \(0 = 0 – 0 + 0 + D \Rightarrow D = 0\)
So \(s = 2t^2 – \frac{t^3}{3} + 6t\) meters
Question 3: The velocity-time graph of a particle is a straight line from (0, -4) to (4, 8). Find:
(a) The acceleration
(b) The displacement from t=0 to t=4
Answer:
(a) Acceleration = slope = \(\frac{8 – (-4)}{4 – 0} = \frac{12}{4} = 3\) m/s²
(b) Displacement = area under v-t graph = area of trapezoid
Area = \(\frac{1}{2} \times (4 + 0) \times (8 + (-4)) = \frac{1}{2} \times 4 \times 4 = 8\) m
Alternatively: \(\frac{1}{2} \times (sum\ of\ parallel\ sides) \times height = \frac{1}{2} \times (-4 + 8) \times 4 = \frac{1}{2} \times 4 \times 4 = 8\) m
Question 4: A particle moves so that its displacement is \(s = 2\sin(3t)\). Find:
(a) Velocity as a function of time
(b) Acceleration as a function of time
Answer:
(a) \(v = \frac{ds}{dt} = 2 \cdot 3\cos(3t) = 6\cos(3t)\) m/s
(b) \(a = \frac{dv}{dt} = 6 \cdot (-3\sin(3t)) = -18\sin(3t)\) m/s²
Question 5: A particle’s acceleration is \(a = 6t\). When \(t=1\), \(v=5\) and \(s=2\). Find the displacement when \(t=3\).
Answer:
First: \(v = \int 6t \, dt = 3t^2 + C\)
Using \(v(1)=5\): \(5 = 3(1)^2 + C \Rightarrow C = 2\)
So \(v = 3t^2 + 2\)
Next: \(s = \int (3t^2 + 2) dt = t^3 + 2t + D\)
Using \(s(1)=2\): \(2 = 1 + 2 + D \Rightarrow D = -1\)
So \(s = t^3 + 2t – 1\)
When \(t=3\): \(s = 27 + 6 – 1 = 32\) m

🎯 Key Concepts Summary

  • Differentiation: \(v = \frac{ds}{dt}\), \(a = \frac{dv}{dt} = \frac{d^2s}{dt^2}\)
  • Integration: \(v = \int a \, dt\), \(s = \int v \, dt\) (with initial conditions)
  • Graphical Relationships:
    • s-t graph: slope = velocity
    • v-t graph: slope = acceleration, area = displacement
    • a-t graph: area = change in velocity
  • Key Moments:
    • At rest: \(v = 0\)
    • Maximum/minimum displacement: \(v = 0\)
    • Constant velocity: \(a = 0\)
    • Direction change: v changes sign
  • Distance vs Displacement:
    • Displacement = net change in position (can be negative)
    • Distance = total path length (always positive)
    • Distance = \(\int |v| \, dt\)
  • CSEC Exam Strategy:
    • Identify what’s given and what’s needed
    • Choose between differentiation or integration
    • For graphs, use slope and area relationships
    • Include initial conditions when integrating
    • Show all steps clearly with formulas
    • Check if answer is reasonable

CSEC Exam Strategy: Displacement-velocity-acceleration questions are common in Paper 2. They often involve: (1) Finding velocity/acceleration from displacement using differentiation, (2) Finding displacement/velocity from acceleration using integration, (3) Interpreting motion graphs, (4) Solving problems with initial conditions. Always state the formula you’re using, show substitution steps, and include units in final answers. For integration problems, don’t forget the constant of integration!

Real-World Applications

Vehicle motion analysis (speed, braking distance)
Projectile motion (balls, rockets, arrows)
Elevator and lift movement
Spring-mass systems (oscillations)
Piston motion in engines
Roller coaster design
Robotic arm movement

Pro Tip: When differentiating or integrating trigonometric functions in motion problems, remember the chain rule: \(\frac{d}{dt}\sin(kt) = k\cos(kt)\) and \(\frac{d}{dt}\cos(kt) = -k\sin(kt)\). For integration: \(\int \sin(kt) dt = -\frac{1}{k}\cos(kt) + C\) and \(\int \cos(kt) dt = \frac{1}{k}\sin(kt) + C\).

Common Mistakes to Avoid: 1. Forgetting to use the chain rule when differentiating trigonometric or composite functions
2. Omitting the constant of integration when finding velocity or displacement from acceleration
3. Confusing distance with displacement (especially when velocity changes sign)
4. Not considering initial conditions when integrating
5. Misinterpreting graphs (e.g., thinking area under s-t graph gives velocity)
6. Forgetting that acceleration can be negative (deceleration)
7. Not including units in final answers

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