Applications of Antidifferentiation to Kinematics
Apply antidifferentiation to motion for Victorian Year 11 Mathematical Methods (VCAA) — recovering velocity and position from acceleration for an object moving in a straight line.
You will learn to antidifferentiate acceleration to find velocity, and velocity to find displacement, using initial conditions to fix each constant, then answer questions about position, speed and when an object is at rest.
Theory
In Year 11 Mathematical Methods (Unit 2), a particle moving in a straight line has position \(x(t)\), velocity \(v(t)=\dfrac{dx}{dt}\) and acceleration \(a(t)=\dfrac{dv}{dt}\). Antidifferentiation runs this chain backwards: integrate the acceleration for the velocity, integrate the velocity for the position. Each step introduces a constant \(+c\) that you pin down with an initial condition. This page shows how to recover \(v(t)\) and \(x(t)\), and how to find when a particle is momentarily at rest.
For motion in a straight line (rectilinear motion), the three quantities are linked by differentiation. Velocity is the rate of change of position and acceleration is the rate of change of velocity:
Antidifferentiation reverses each arrow. Given the acceleration, antidifferentiate to recover the velocity; given the velocity, antidifferentiate to recover the position:
Every antidifferentiation brings an arbitrary constant \(+c\). An initial condition — the velocity or position at \(t=0\) — fixes that constant so the answer describes one particular motion.
Antidifferentiate up the chain, adding a constant each time:
The basic power rule used throughout is
Useful readings from the motion functions:
- At rest: solve \(v(t)=0\).
- Velocity / acceleration at a time: substitute \(t\) into \(v(t)\) or \(a(t)\).
- A negative \(v(t)\) means motion in the negative direction; a negative \(x(t)\) means the particle is on the negative side of its start.
Recovering velocity and position by antidifferentiation
- Start from what you are given — the acceleration \(a(t)\) or the velocity \(v(t)\).
- Antidifferentiate term by term to get the next function, remembering the constant \(+c\).
- Apply the initial condition (\(v(0)\), \(x(0)\), or ‘starts from rest’ meaning \(v(0)=0\)) and solve for \(c\).
- Repeat if needed: antidifferentiate the velocity to reach the position, applying a second initial condition for its constant.
- Answer the question: substitute the required time, or solve \(v(t)=0\) for an ‘at rest’ time. Attach the correct metric unit (\(\text{m}\), \(\text{m/s}\), \(\text{m/s}^2\)).
Antidifferentiate the acceleration to get the velocity:
| \(v(t)\) | \(=\) | \(\int 6\,dt\) |
| \(=\) | \(6t+c\) |
Apply the initial condition \(v(0)=4\):
| \(4\) | \(=\) | \(6(0)+c\) |
| \(c\) | \(=\) | \(4\) |
| \(v(t)\) | \(=\) | \(6t+4\) |
Evaluate at \(t=3\):
| \(v(3)\) | \(=\) | \(6(3)+4\) |
| \(=\) | \(22\) |
The velocity is \(22\ \text{m/s}\).
Antidifferentiate the velocity to get the position:
| \(x(t)\) | \(=\) | \(\int (4t+3)\,dt\) |
| \(=\) | \(2t^2+3t+c\) |
Apply the initial condition \(x(0)=5\):
| \(5\) | \(=\) | \(2(0)^2+3(0)+c\) |
| \(c\) | \(=\) | \(5\) |
| \(x(t)\) | \(=\) | \(2t^2+3t+5\) |
Evaluate at \(t=2\):
| \(x(2)\) | \(=\) | \(2(2)^2+3(2)+5\) |
| \(=\) | \(8+6+5\) | |
| \(=\) | \(19\) |
The position is \(19\ \text{m}\).
Antidifferentiate the acceleration to get the velocity:
| \(v(t)\) | \(=\) | \(\int (-10)\,dt\) |
| \(=\) | \(-10t+c\) |
Apply the initial condition \(v(0)=25\):
| \(25\) | \(=\) | \(-10(0)+c\) |
| \(c\) | \(=\) | \(25\) |
| \(v(t)\) | \(=\) | \(-10t+25\) |
At rest means \(v(t)=0\):
| \(-10t+25\) | \(=\) | \(0\) |
| \(10t\) | \(=\) | \(25\) |
| \(t\) | \(=\) | \(2.5\) |
The ball is momentarily at rest when \(t=2.5\ \text{s}\).
Velocity — antidifferentiate the acceleration:
| \(v(t)\) | \(=\) | \(\int (6t-24)\,dt\) |
| \(=\) | \(3t^2-24t+c\) |
Apply the initial condition \(v(0)=36\):
| \(36\) | \(=\) | \(3(0)^2-24(0)+c\) |
| \(c\) | \(=\) | \(36\) |
| \(v(t)\) | \(=\) | \(3t^2-24t+36\) |
(i) Evaluate the velocity at \(t=3\):
| \(v(3)\) | \(=\) | \(3(3)^2-24(3)+36\) |
| \(=\) | \(27-72+36\) | |
| \(=\) | \(-9\) |
The velocity is \(-9\ \text{m/s}\).
(ii) At rest means \(v(t)=0\); solve the quadratic:
| \(3t^2-24t+36\) | \(=\) | \(0\) |
| \(3(t^2-8t+12)\) | \(=\) | \(0\) |
| \(3(t-2)(t-6)\) | \(=\) | \(0\) |
| \(t\) | \(=\) | \(2 \text{ or } 6\) |
The earliest rest time is \(t=2\ \text{s}\).
(iii) Position — antidifferentiate the velocity:
| \(x(t)\) | \(=\) | \(\int (3t^2-24t+36)\,dt\) |
| \(=\) | \(t^3-12t^2+36t+c\) |
Apply \(x(0)=0\), then evaluate at \(t=2\):
| \(c\) | \(=\) | \(0\) |
| \(x(2)\) | \(=\) | \((2)^3-12(2)^2+36(2)\) |
| \(=\) | \(8-48+72\) | |
| \(=\) | \(32\) |
(i) \(v=-9\ \text{m/s}\)\quad (ii) \(t=2\ \text{s}\)\quad (iii) \(x=32\ \text{m}\).
Common pitfalls
Frequently asked questions
How do you find velocity from acceleration?
Antidifferentiate the acceleration, \(v(t)=\int a(t)\,dt\), then use an initial velocity such as \(v(0)\) to find the constant \(c\).
How do you find position from velocity?
Antidifferentiate the velocity, \(x(t)=\int v(t)\,dt\), then apply an initial position such as \(x(0)\) to find the constant \(c\).
What does ‘momentarily at rest’ mean?
The velocity is zero at that instant. Solve \(v(t)=0\) for \(t\); the particle may still have non-zero acceleration and start moving again.
Why do we add a constant when antidifferentiating?
Antidifferentiation reverses differentiation, and any constant differentiates to zero, so infinitely many functions share the same derivative. The \(+c\) is fixed by an initial condition.
What does ‘starts from rest’ tell me?
It gives the initial condition \(v(0)=0\), which you substitute to find the constant when antidifferentiating the acceleration.