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Year 11 Methods (Unit 1 & 2) Applications Of Differentiation Of Polynomials

Applications Of Differentiation To Kinematics

20 practice questions 1 video lesson Theory + worked examples

Master the calculus of motion for Queensland Year 11 Mathematical Methods (QCAA). On a displacement-time graph for a particle in a straight line, velocity is the slope of the tangent — the instantaneous rate of change of displacement.

You will learn to find velocity and acceleration by differentiating a displacement function, identify when a particle is at rest, read its direction of motion, and locate greatest displacement with a sign test.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), for a particle moving in a straight line the velocity is the derivative of the displacement, \(v=\dfrac{dx}{dt}\), and the acceleration is the derivative of the velocity, \(a=\dfrac{dv}{dt}\). This page shows how to differentiate a polynomial position function to find velocity and acceleration, evaluate them at a time, find when a particle is at rest, read its direction, and find greatest velocity or maximum displacement with a first-derivative sign test.

For straight-line motion, displacement \(x(t)\) gives the particle's position. Its rate of change is the velocity, \(v=\dfrac{dx}{dt}\) — the gradient of the displacement-time graph. The rate of change of velocity is the acceleration, \(a=\dfrac{dv}{dt}\).

The sign of the velocity gives the direction of motion: positive moves in the positive direction, negative in the negative direction. The particle is at rest when \(v=0\). Initial values are found by putting \(t=0\).

Differentiate down the chain. Position \(\xrightarrow{\ d/dt\ }\) velocity \(\xrightarrow{\ d/dt\ }\) acceleration. In Year 11 you only differentiate — you never integrate velocity back to position.
Velocity as the slope of the displacement-time graphDisplacement-time curve with the tangent at t=4; its gradient is the velocity there. x y v = slope
Velocity is the gradient of the displacement-time graph, \(v=\dfrac{dx}{dt}\).
Velocity-time graphVelocity-time curve crossing zero at t=4 where the particle is momentarily at rest. x y v = 0
On a velocity-time graph, \(v=0\) marks where the particle is momentarily at rest.

Velocity is the derivative of displacement:

\[v=\dfrac{dx}{dt}\]
v=dxdt

Acceleration is the derivative of velocity:

\[a=\dfrac{dv}{dt}\]
a=dvdt
Key conditions: at rest \(\Rightarrow v=0\); initial values \(\Rightarrow t=0\); greatest velocity or maximum displacement \(\Rightarrow\) the quantity's derivative is \(0\), confirmed by a first-derivative sign test.

How to solve a kinematics problem by differentiation

  1. Differentiate: from \(x(t)\) find \(v=\dfrac{dx}{dt}\); differentiate again for \(a=\dfrac{dv}{dt}\).
  2. Substitute or solve: put in a time for a value, use \(t=0\) for initial values, or set \(v=0\) to find when the particle is at rest.
  3. Interpret: read direction from the sign of \(v\); for greatest velocity or maximum displacement, set the relevant derivative to \(0\) and confirm with a sign test.
Example 1 — Velocity from displacement
A particle moves so that \(x(t)=t^3-6t^2+9t\) metres after \(t\) seconds. Find \(v(t)\) and the velocity at \(t=4\).
Solution

Velocity — differentiate \(x(t)\):

\(v=\dfrac{dx}{dt}\)\(=\)\(3t^2-12t+9\)

Value at \(t=4\):

\(v(4)\)\(=\)\(3(4)^2-12(4)+9\)
\(=\)\(48-48+9\)
\(=\)\(9\)

\(v(t)=3t^2-12t+9\); at \(t=4\) s the velocity is \(9\) m/s.

Velocity from a displacement-time graphDisplacement-time curve with the tangent of gradient 9 at t=4 seconds. x y
v=9 m/s
Example 2 — Acceleration and initial value
For the same motion \(x(t)=t^3-6t^2+9t\), find the acceleration \(a(t)\), its value at \(t=3\), and the initial acceleration.
Solution

Acceleration — differentiate \(v(t)=3t^2-12t+9\):

\(a=\dfrac{dv}{dt}\)\(=\)\(6t-12\)

Value at \(t=3\):

\(a(3)\)\(=\)\(6(3)-12\)
\(=\)\(6\)

Initial acceleration — put \(t=0\):

\(a(0)\)\(=\)\(6(0)-12\)
\(=\)\(-12\)

\(a(t)=6t-12\); \(a(3)=6\) m/s\(^2\), and the initial acceleration is \(-12\) m/s\(^2\).

Acceleration as the slope of the velocity-time graphVelocity-time curve with the tangent of gradient 6 at t=3, the acceleration. x y
a(3)=6
Example 3 — At rest and direction of motion
A particle has velocity \(v(t)=3t^2-12t+9\) m/s. Find when it is at rest, and its direction of motion at \(t=2\).
Solution

At rest — set \(v=0\):

\(3t^2-12t+9\)\(=\)\(0\)
\(3(t-1)(t-3)\)\(=\)\(0\)
\(t\)\(=\)\(1 \ \text{or}\ 3\)

Direction at \(t=2\) — find the sign of \(v\):

\(v(2)\)\(=\)\(3(2-1)(2-3)\)
\(=\)\(3(1)(-1)\)
\(=\)\(-3\)

\(v(2)<0\), so the particle is moving in the negative direction.

At rest at \(t=1\) s and \(t=3\) s; at \(t=2\) s it moves in the negative direction (\(v=-3\) m/s).

Times when the particle is at restVelocity-time curve crossing zero at t=1 and t=3 where the particle is at rest. x y
t=1 or t=3
Example 4 — Maximum displacement by the sign test
A particle's displacement is \(x(t)=6t^2-t^3\) metres for \(t\ge 0\). Find its maximum displacement.
Solution

Velocity — differentiate, then set \(v=0\):

\(v=\dfrac{dx}{dt}\)\(=\)\(12t-3t^2\)
\(3t(4-t)\)\(=\)\(0\)
\(t\)\(=\)\(0 \ \text{or}\ 4\)

Sign test around \(t=4\):

\(v(3)\)\(=\)\(3(3)(4-3)=9>0\)
\(v(5)\)\(=\)\(3(5)(4-5)=-15<0\)

\(+\rightarrow-\), so displacement is greatest (a maximum) at \(t=4\) s.

Maximum displacement — substitute \(t=4\):

\(x(4)\)\(=\)\(6(4)^2-(4)^3\)
\(=\)\(96-64\)
\(=\)\(32\)

The maximum displacement is \(32\) m, at \(t=4\) s.

Maximum displacement where velocity is zeroVelocity-time curve for the motion, crossing zero at t=4 where displacement is greatest. x y
x=32 m

Common pitfalls

Confusing position with velocity. \(x(t)\) is where the particle is; the velocity is its derivative \(\dfrac{dx}{dt}\). Differentiate before reading a speed.
Trying to integrate. In Year 11 you only differentiate — going from velocity back to displacement by integration is Year 12.
Forgetting the units. Velocity is in m/s and acceleration in m/s\(^2\); keep the sign, as it carries the direction.
Skipping the sign test. For greatest velocity or maximum displacement, setting the derivative to \(0\) is not enough — confirm the maximum with a first-derivative sign test.

Frequently asked questions

How do you find velocity from displacement?

Differentiate the displacement with respect to time: \(v=\dfrac{dx}{dt}\). The velocity is the gradient of the displacement-time graph.

How do you find acceleration?

Differentiate the velocity with respect to time: \(a=\dfrac{dv}{dt}\), which is the second derivative of displacement.

When is a particle at rest?

When its velocity is zero. Set \(v(t)=0\) and solve for \(t\).

How do I find the initial velocity or acceleration?

Substitute \(t=0\) into \(v(t)\) or \(a(t)\); \"initial\" always means the value at \(t=0\).

What does the sign of the velocity mean?

A positive velocity means the particle moves in the positive direction; a negative velocity means it moves in the negative direction.

Do you integrate in Year 11 kinematics?

No. Year 11 Methods uses differentiation only — you never integrate velocity back to displacement, and maxima are confirmed by the first-derivative sign test.

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