Motion in a straight line (vector calculus)
Use the calculus of vector functions to describe motion in a straight line for Year 12 Specialist Mathematics in Queensland (QCAA). Writing a particle’s position as a vector function of time, you differentiate to find velocity and acceleration and integrate, using initial conditions, to work back the other way.
You will learn to handle both constant and variable acceleration, find when a particle is at rest or changes direction, and tell displacement apart from distance travelled — the rectilinear-motion core of Unit 3 vector calculus.
Theory
Motion in a straight line (vector calculus) uses the calculus of vector functions to link a particle’s position, velocity and acceleration in Year 12 Specialist Mathematics (QCAA, Queensland). Writing the position as \(r(t)=x(t)\,i\), you differentiate to move from position to velocity to acceleration, and integrate (using initial conditions) to move back — for both constant and variable acceleration.
A particle moving along a line has a position vector \(r(t)=x(t)\,i\), where \(x(t)\) is its signed displacement from the origin at time \(t\) and \(i\) is the unit vector along the line. Its velocity is the rate of change of position, and its acceleration is the rate of change of velocity.
Velocity is the derivative of position, \(v(t)=\dot r(t)=\dfrac{dr}{dt}\), and acceleration is the derivative of velocity, \(a(t)=\dot v(t)=\ddot r(t)=\dfrac{d^2r}{dt^2}\). Differentiating with respect to time moves you down the chain: position \(\to\) velocity \(\to\) acceleration.
Going the other way, you integrate: \(v(t)=\int a\,dt\) and \(r(t)=\int v\,dt\). Each integration introduces a constant vector, which you fix from an initial condition — usually the velocity or position when \(t=0\).
The particle is momentarily at rest when \(v(t)=0\); this is also where it may change direction. Displacement over an interval is the net change \(r(t_2)-r(t_1)\), while distance travelled adds the length of each leg between direction changes.
For a particle with position \(r(t)=x(t)\,i\), velocity and acceleration are successive time derivatives:
Reversing the chain, integrate and fix each constant from an initial condition:
Over a time interval, the displacement is the net change of position, and the distance travelled sums the legs between rest points:
Working between position, velocity and acceleration
- Identify what you are given (position \(r\), velocity \(v\), or acceleration \(a\)) and what is asked.
- Differentiate to go down the chain (\(r\to v\to a\)); integrate to go up (\(a\to v\to r\)), writing \(+c\) each time.
- Apply the initial conditions (the values at \(t=0\)) to solve for each constant of integration.
- Answer the question: substitute a time for an instant value, set \(v=0\) for rest, take \(r(t_2)-r(t_1)\) for displacement, or add the legs between rest points for distance.
Velocity is the first derivative of position, \(v=\dot r\):
| \(v(t)\) | \(=\) | \(\dfrac{d}{dt}(t^3-6t^2+9t)\,i\) |
| \(=\) | \((3t^2-12t+9)\,i\) |
Acceleration is the derivative of velocity, \(a=\dot v\):
| \(a(t)\) | \(=\) | \(\dfrac{d}{dt}(3t^2-12t+9)\,i\) |
| \(=\) | \((6t-12)\,i\) |
Substitute \(t=0\) into the velocity:
| \(v(0)\) | \(=\) | \((3(0)^2-12(0)+9)\,i\) |
| \(=\) | \(9i\) |
\(v(t)=(3t^2-12t+9)i\), \(a(t)=(6t-12)i\), and \(v(0)=9i\) m/s.
Integrate the acceleration; the initial velocity fixes the constant:
| \(v(t)\) | \(=\) | \(\int (6t+2)\,dt\;i\) |
| \(=\) | \((3t^2+2t+c)\,i\) | |
| \(v(0)\) | \(=\) | \(1i \Rightarrow c=1\) |
| \(v(t)\) | \(=\) | \((3t^2+2t+1)\,i\) |
Integrate the velocity; the starting position fixes the new constant:
| \(r(t)\) | \(=\) | \(\int (3t^2+2t+1)\,dt\;i\) |
| \(=\) | \((t^3+t^2+t+c)\,i\) | |
| \(r(0)\) | \(=\) | \(0 \Rightarrow c=0\) |
| \(r(t)\) | \(=\) | \((t^3+t^2+t)\,i\) |
Substitute \(t=2\) into the position:
| \(r(2)\) | \(=\) | \(((2)^3+(2)^2+2)\,i\) |
| \(=\) | \((8+4+2)\,i\) | |
| \(=\) | \(14i\) |
\(v(t)=(3t^2+2t+1)i\), \(r(t)=(t^3+t^2+t)i\), and \(r(2)=14i\) m.
Differentiate for the velocity, then set it to zero for the rest time:
| \(v(t)\) | \(=\) | \((2t-6)\,i\) |
| \(2t-6\) | \(=\) | \(0\) |
| \(t\) | \(=\) | \(3\) |
Displacement is the net change \(r(5)-r(0)\):
| \(r(0)\) | \(=\) | \(((0)^2-6(0)+5)\,i=5i\) |
| \(r(5)\) | \(=\) | \(((5)^2-6(5)+5)\,i=0i\) |
| \(r(5)-r(0)\) | \(=\) | \(0i-5i=-5i\) |
It turns at \(t=3\); add the length of each leg for the distance:
| \(r(3)\) | \(=\) | \(((3)^2-6(3)+5)\,i=-4i\) |
| \(\text{distance}\) | \(=\) | \(|{-4}-5|+|0-({-4})|\) |
| \(=\) | \(9+4\) | |
| \(=\) | \(13\) |
At rest at \(t=3\) s; displacement \(-5i\) m; distance \(13\) m.
Integrate the acceleration; \(v(0)=0\) gives \(c=0\):
| \(v(t)\) | \(=\) | \(\int (6-6t)\,dt\;i\) |
| \(=\) | \((6t-3t^2+c)\,i\) | |
| \(=\) | \((6t-3t^2)\,i\) |
Integrate the velocity; \(r(0)=0\) gives \(c=0\):
| \(r(t)\) | \(=\) | \(\int (6t-3t^2)\,dt\;i\) |
| \(=\) | \((3t^2-t^3)\,i\) |
The displacement is greatest where the velocity is zero:
| \(6t-3t^2\) | \(=\) | \(0\) |
| \(3t(2-t)\) | \(=\) | \(0\) |
| \(t\) | \(=\) | \(2 \;\; (t>0)\) |
| \(r(2)\) | \(=\) | \((3(2)^2-(2)^3)\,i\) |
| \(=\) | \((12-8)\,i=4i\) |
\(v(t)=(6t-3t^2)i\), \(r(t)=(3t^2-t^3)i\); greatest displacement \(4\) m at \(t=2\).
Common pitfalls
Frequently asked questions
How do you find velocity and acceleration from a position vector?
Differentiate with respect to time. Velocity is \(v=\dot r=\dfrac{dr}{dt}\) and acceleration is \(a=\dot v=\ddot r=\dfrac{d^2r}{dt^2}\), so acceleration is the second derivative of position.
How do you get velocity and position from acceleration?
Integrate: \(v=\int a\,dt\) and \(r=\int v\,dt\). Each integration adds a constant, which you find from the initial velocity or position (the values when \(t=0\)).
When is a particle momentarily at rest?
When its velocity is zero. Solve \(v(t)=0\); these times are also where the particle can change direction.
What is the difference between displacement and distance travelled?
Displacement is the net change in position \(r(t_2)-r(t_1)\). Distance travelled adds the length of each leg between direction changes, so it is never negative and is often larger.
How do you handle variable acceleration such as \(a=6-6t\)?
The same way as constant acceleration: integrate term by term to get \(v\), then \(r\), using the initial conditions for the constants. Only the algebra changes, not the method.
Why does the answer keep the unit vector \(i\)?
Because position, velocity and acceleration are vectors along the line. The \(i\) records the direction, and its sign tells you which way the particle is moving.