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.
Differentiation moves downward from displacement to acceleration; integration moves upward from acceleration to displacement.
\(s \xrightarrow{\dfrac{d}{dt}} v \xrightarrow{\dfrac{d}{dt}} a\)
Velocity is the rate of change of displacement; acceleration is the rate of change of velocity.
\(a \xrightarrow{\int dt} v \xrightarrow{\int dt} s\)
Integrate acceleration to obtain velocity, then integrate velocity to obtain displacement.
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.
Use integration to move from acceleration to velocity and from velocity to displacement.
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 \(c\).
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 \(c\).
Perform two consecutive integrations: first \(v=\int a\,dt\), then \(s=\int v\,dt\). Each step produces its own constant of integration.
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.
A particle is instantaneously at rest when its instantaneous velocity is zero, \(v=0\). This often marks a turning point in linear motion.
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 \(v=0\).
Velocity from acceleration
Find \(c\) using initial velocity at \(t=0\).
Displacement from velocity
Find \(c\) using initial position at \(t=0\).
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 \(t=6\text{ s}\).
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.
Objective: Test your understanding of integration in kinematics, velocity, displacement and distance.
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?
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.
Drag each term to the correct meaning. On a phone, tap a term first, then tap the matching box.
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