8.3 Integration in Kinematics of Linear Motion
Integration reverses differentiation in linear motion. Given acceleration \(a\), integrate once to obtain velocity \(v=\int a\,dt\), and integrate velocity to obtain displacement \(s=\int v\,dt\). Initial conditions determine the constants of integration.
Learning Objectives
- Determine and interpret instantaneous velocity from an acceleration function by integrating with respect to time.
- Determine the constant of integration in the velocity function using given initial conditions.
- Find maximum or minimum velocity by applying \(a=0\) and testing the second derivative.
- Determine and interpret instantaneous displacement from velocity or acceleration functions.
- Calculate distance travelled in the \(n^{\text{th}}\) second and total distance travelled using number line analysis.
Integration and Differentiation Workflow in Linear Motion
Differentiation moves downward from displacement to acceleration; integration moves upward from acceleration to displacement.
Differentiation Path
\(s \xrightarrow{\dfrac{d}{dt}} v \xrightarrow{\dfrac{d}{dt}} a\)
Velocity is the rate of change of displacement; acceleration is the rate of change of velocity.
Integration Path
\(a \xrightarrow{\int dt} v \xrightarrow{\int dt} s\)
Integrate acceleration to obtain velocity, then integrate velocity to obtain displacement.
Number Line Method for Total Distance
Mark displacement at \(t=0\).
Find times when \(v=0\) and mark their displacements.
Mark displacement at the final time \(t=n\).
Sum the absolute distances between all turning points and endpoints.
Short Explanation
Use integration to move from acceleration to velocity and from velocity to displacement.
1. Velocity from Acceleration
Given \(a=\dfrac{dv}{dt}\), integrate to obtain \(v=\int a\,dt\). Use an initial condition such as \(v=u\) at \(t=0\) to find the constant \(c\).
2. Displacement from Velocity
Given \(v=\dfrac{ds}{dt}\), integrate to obtain \(s=\int v\,dt\). Use an initial position such as \(s=0\) at \(t=0\) to find the constant \(c\).
3. Displacement from Acceleration
Perform two consecutive integrations: first \(v=\int a\,dt\), then \(s=\int v\,dt\). Each step produces its own constant of integration.
4. Maximum and Minimum Velocity
Maximum or minimum velocity occurs when \(\dfrac{dv}{dt}=a=0\). Use \(\dfrac{d^2v}{dt^2}\) to determine the nature: positive means minimum, negative means maximum.
5. Instantaneous Rest
A particle is instantaneously at rest when its instantaneous velocity is zero, \(v=0\). This often marks a turning point in linear motion.
6. Distance vs Displacement
Displacement is a vector and can be negative. Distance is a scalar and is always positive or zero. Total distance requires checking for reversals where \(v=0\).
| Condition | Mathematical Test | Interpretation |
|---|---|---|
| Maximum velocity | \(a=0\) and \(\dfrac{d^2v}{dt^2}<0\) | Velocity reaches its highest value. |
| Minimum velocity | \(a=0\) and \(\dfrac{d^2v}{dt^2}>0\) | Velocity reaches its lowest value. |
| Instantaneous rest | \(v=0\) | The particle stops momentarily and may reverse direction. |
| Distance in \(n^{\text{th}}\) second | \(|s_n-s_{n-1}|\) | Distance covered between \(t=n-1\) and \(t=n\). |
Velocity from acceleration
\(v=\int a\,dt\)Find \(c\) using initial velocity at \(t=0\).
Displacement from velocity
\(s=\int v\,dt\)Find \(c\) using initial position at \(t=0\).
Try to Answer First
Answer in your mind, then press "Check Answer".
A particle passes a fixed point \(O\) with an initial velocity of \(10\text{ ms}^{-1}\). Its acceleration is given by \(a=4t-8\). Find the velocity function \(v\) in terms of \(t\), and find the minimum velocity of the particle.
A particle moves along a straight line such that its velocity is \(v=12+4t-t^2\). Determine the displacement function \(s\) if the particle starts at the origin \(O\), and calculate its displacement when the particle is instantaneously at rest for \(t>0\).
Explain how to calculate the distance travelled by a particle in the \(7^{\text{th}}\) second if its displacement function is \(s=12t+2t^2-\dfrac{1}{3}t^3\) and it only changes direction at \(t=6\text{ s}\).
Common Mistakes
- Forgetting the constant of integration \(c\) when writing \(v=\int a\,dt\) or \(s=\int v\,dt\).
- Assuming \(c=0\) automatically without substituting the given initial conditions.
- Confusing displacement with total distance travelled, ignoring reversals in motion where \(v=0\).
- Confusing distance in the \(n^{\text{th}}\) second with total distance in \(n\) seconds; use \(|s_n-s_{n-1}|\), not \(s_n\).
- Accepting negative time values when solving quadratic equations in \(t\), overlooking that time \(t\geq0\).
Concept Misunderstandings
When acceleration is zero, the particle stops moving.
When \(a=0\), velocity is stationary — at a maximum, minimum, or uniform velocity. The particle stops only when \(v=0\).
Distance travelled can be negative.
Displacement can be negative because it is a vector. Distance is a scalar and is always positive or zero.
\(\dfrac{d^2v}{dt^2}<0\) means the velocity is minimum.
A negative second derivative indicates a maximum stationary value; \(\dfrac{d^2v}{dt^2}>0\) indicates a minimum.
Summary
- Velocity from acceleration: \(v=\int a\,dt\). Find \(c\) using initial velocity at \(t=0\).
- Displacement from velocity: \(s=\int v\,dt\). Find \(c\) using initial position at \(t=0\).
- Displacement from acceleration requires two consecutive integrations: \(v=\int a\,dt\), then \(s=\int v\,dt\).
- Turning point / instantaneous rest occurs when \(v=0\).
- Maximum or minimum velocity occurs when \(a=0\); test the sign of \(\dfrac{d^2v}{dt^2}\).
- Distance in the \(n^{\text{th}}\) second is \(|s_n-s_{n-1}|\).
- Total distance is the sum of absolute displacement changes across intervals bounded by turning points.
Short Activity
Objective: Test your understanding of integration in kinematics, velocity, displacement and distance.
A. Objective Quiz
1 A particle moves in a straight line with an acceleration given by \(a=6-2t\text{ ms}^{-2}\). If the initial velocity of the particle is \(5\text{ ms}^{-1}\), what is the maximum velocity of the particle?
2 The displacement of a particle moving in a straight line is given by \(s=4t^2-t^3\). What formula represents the distance travelled by the particle in the \(4^{\text{th}}\) second?
B. Fill in the Blanks
3 When the velocity function \(v\) is given as a function of time \(t\), the displacement function \(s\) can be determined using the formula \(s=\int\) ________ \(dt\).
4 A particle moving linearly is instantaneously at rest when its instantaneous ________ is equal to zero.
C. Matching / Drag and Drop
Drag each term to the correct meaning. On a phone, tap a term first, then tap the matching box.
