Motion in a straight line (dynamics)
Model motion in a straight line for Year 12 Specialist Mathematics in Queensland (QCAA). By differentiating a particle’s displacement you find its velocity and acceleration, and by integrating — using initial conditions — you work back the other way, for both constant and non-constant acceleration.
You will learn the suvat rules for constant acceleration and vertical motion under gravity, use \(a=v\dfrac{dv}{dx}\) when acceleration depends on position, find when a particle is at rest, and tell displacement apart from distance travelled — the core of Unit 4 modelling motion.
Theory
Motion in a straight line (dynamics) models a particle’s displacement, velocity and acceleration along a line in Year 12 Specialist Mathematics (QCAA, Queensland). You differentiate to move from \(x\) to \(v\) to \(a\), and integrate (using initial conditions) to move back — for constant acceleration (the suvat rules) and for acceleration that varies with time or position.
A particle moving along a line has a signed displacement \(x\) from a fixed origin at time \(t\). Its velocity \(v\) is the rate of change of displacement, and its acceleration \(a\) is the rate of change of velocity. The speed is \(|v|\), the size of the velocity regardless of direction.
Differentiate to move down the chain: \(v=\dfrac{dx}{dt}\) and \(a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}\). When the acceleration is given in terms of position rather than time, use \(a=v\dfrac{dv}{dx}=\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\).
Integrate to move back up: \(v=\int a\,dt\) and \(x=\int v\,dt\). Each integration adds a constant, fixed from an initial condition (the value at \(t=0\)). When the acceleration is constant, this leads to the suvat equations, such as \(v=u+at\) and \(v^2=u^2+2as\).
The particle is momentarily at rest when \(v=0\); this is also where it may change direction. Displacement over an interval is the net change \(x(t_2)-x(t_1)\), while distance travelled adds the length of each leg between direction changes, so it can be larger.
Displacement, velocity and acceleration are linked by differentiation (down) and integration (up):
Reverse the chain by integrating, fixing each constant from an initial condition:
For constant acceleration only, the suvat equations relate initial velocity \(u\), final velocity \(v\), acceleration \(a\), displacement \(s\) and time \(t\):
Working between displacement, velocity and acceleration
- Identify what you are given (\(x\), \(v\) or \(a\)) and whether the acceleration is constant, varies with time, or varies with position.
- Differentiate to go down the chain (\(x\to v\to a\)); integrate to go up (\(a\to v\to x\)), writing \(+c\) each time. If \(a\) depends on position, use \(a=v\dfrac{dv}{dx}\).
- Apply the initial conditions (the values at \(t=0\)) to solve for every constant of integration, or use the suvat equations when \(a\) is constant.
- Answer the question: substitute a time for an instant value, set \(v=0\) for rest, take \(x(t_2)-x(t_1)\) for displacement, or add the legs between rest points for distance.
Velocity is the derivative of displacement, \(v=\dfrac{dx}{dt}\):
| \(v\) | \(=\) | \(\dfrac{d}{dt}(t^3-3t^2+2t)\) |
| \(=\) | \(3t^2-6t+2\) |
Acceleration is the derivative of velocity, \(a=\dfrac{dv}{dt}\):
| \(a\) | \(=\) | \(\dfrac{d}{dt}(3t^2-6t+2)\) |
| \(=\) | \(6t-6\) |
Substitute \(t=1\) into the velocity:
| \(v(1)\) | \(=\) | \(3(1)^2-6(1)+2\) |
| \(=\) | \(3-6+2\) | |
| \(=\) | \(-1\) |
\(v=3t^2-6t+2\) m/s, \(a=6t-6\) m/s\(^2\), and \(v(1)=-1\) m/s.
Integrate the acceleration; the initial velocity \(v(0)=4\) fixes the constant:
| \(v\) | \(=\) | \(\int (6t-2)\,dt\) |
| \(=\) | \(3t^2-2t+c\) | |
| \(v(0)=4\) | \(\Rightarrow\) | \(c=4\) |
| \(v\) | \(=\) | \(3t^2-2t+4\) |
Integrate the velocity; the starting position \(x(0)=1\) fixes the new constant:
| \(x\) | \(=\) | \(\int (3t^2-2t+4)\,dt\) |
| \(=\) | \(t^3-t^2+4t+c\) | |
| \(x(0)=1\) | \(\Rightarrow\) | \(c=1\) |
| \(x\) | \(=\) | \(t^3-t^2+4t+1\) |
Substitute \(t=2\) into the displacement:
| \(x(2)\) | \(=\) | \((2)^3-(2)^2+4(2)+1\) |
| \(=\) | \(8-4+8+1\) | |
| \(=\) | \(13\) |
\(v=3t^2-2t+4\) m/s, \(x=t^3-t^2+4t+1\) m, and \(x(2)=13\) m.
Differentiate for the velocity, then set it to zero for the rest time:
| \(v\) | \(=\) | \(2t-6\) |
| \(2t-6\) | \(=\) | \(0\) |
| \(t\) | \(=\) | \(3\) |
Displacement is the net change \(x(6)-x(0)\):
| \(x(0)\) | \(=\) | \((0)^2-6(0)=0\) |
| \(x(6)\) | \(=\) | \((6)^2-6(6)=0\) |
| \(x(6)-x(0)\) | \(=\) | \(0-0=0\) |
It turns at \(t=3\); add the length of each leg for the distance:
| \(x(3)\) | \(=\) | \((3)^2-6(3)=-9\) |
| \(\text{distance}\) | \(=\) | \(|{-9}-0|+|0-({-9})|\) |
| \(=\) | \(9+9\) | |
| \(=\) | \(18\) |
At rest at \(t=3\) s; displacement \(0\) m; distance travelled \(18\) m.
The velocity is given in terms of \(x\), so use \(a=v\dfrac{dv}{dx}\):
| \(v\) | \(=\) | \(4x+1\) |
| \(\dfrac{dv}{dx}\) | \(=\) | \(4\) |
| \(a\) | \(=\) | \(v\dfrac{dv}{dx}\) |
| \(=\) | \((4x+1)(4)\) | |
| \(=\) | \(16x+4\) |
Substitute \(x=2\) into the acceleration:
| \(a(2)\) | \(=\) | \(16(2)+4\) |
| \(=\) | \(32+4\) | |
| \(=\) | \(36\) |
\(a=16x+4\), so the acceleration when \(x=2\) is \(36\) m/s\(^2\).
Common pitfalls
Frequently asked questions
How do you find velocity and acceleration from displacement?
Differentiate with respect to time. Velocity is \(v=\dfrac{dx}{dt}\) and acceleration is \(a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}\), so acceleration is the second derivative of displacement.
How do you find velocity and displacement from acceleration?
Integrate: \(v=\int a\,dt\) and \(x=\int v\,dt\). Each integration adds a constant, which you find from the initial velocity or displacement (the values when \(t=0\)).
When is a particle at rest or changing direction?
When its velocity is zero. Solve \(v=0\); these times are where the particle is momentarily at rest and may reverse direction.
What is the difference between distance and displacement?
Displacement is the net change in position \(x(t_2)-x(t_1)\). Distance travelled adds the length of each leg between direction changes, so it is never negative and is often larger.
What are the suvat equations and when can you use them?
For constant acceleration, \(v=u+at\), \(s=ut+\tfrac12 at^2\) and \(v^2=u^2+2as\). They apply only when the acceleration is constant; for variable acceleration you must integrate.
What does \(a=v\dfrac{dv}{dx}\) mean and when do you use it?
It is another form of acceleration, equal to \(\dfrac{d}{dx}\!\left(\tfrac12 v^2\right)\). Use it when velocity or acceleration is given in terms of position \(x\) rather than time \(t\).