How do you find the maximum displacement or maximum velocity?

Maximum displacement occurs at a stationary point of the displacement, where the velocity is zero; maximum or minimum velocity occurs where the acceleration is zero. Find that time, confirm it is a maximum or minimum with a second derivative or sign test, then evaluate the quantity, and also check the endpoints of any given interval.

How do you find the maximum displacement or maximum velocity?

Maximum displacement is at a stationary point where v equals zero; maximum or minimum velocity is where a equals zero. Confirm with a second derivative test and check the interval endpoints.

Year 12 Maths - Methods (Unit 3 & Unit 4) Applications of differentiation

Motion in a straight line

Q: How do you find velocity and acceleration from displacement?

Velocity is the derivative of displacement, v equals dx by dt, and acceleration is the derivative of velocity, a equals dv by dt, the second derivative of displacement. Differentiate once for velocity and twice for acceleration, then substitute the required time.

Q: When is a particle momentarily at rest?

When its velocity is zero. Differentiate the displacement, set v equal to zero and solve for the time. The particle stops there before it changes direction.

Q: How do you tell which direction a particle is moving?

By the sign of the velocity: positive means the positive direction, negative means the negative direction, and zero means momentarily at rest.

Q: What is the difference between speeding up and slowing down?

Speeding up means v and a have the same sign (v times a positive); slowing down means they have opposite signs (v times a negative).

Q: What is the difference between distance and displacement?

Displacement is the signed change in position; distance is the total path length and is never negative. When the particle reverses, add the distances between successive rest positions.

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Theory

Motion in a straight line is analysed with calculus. Differentiate the displacement \(x(t)\) once for the velocity \(v=\dfrac{dx}{dt}\) and again for the acceleration \(a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}\). The sign of \(v\) gives the direction, \(v=0\) marks where the particle is momentarily at rest, and the sign of \(v\,a\) tells you whether it is speeding up or slowing down.

For a particle moving on a straight line, its displacement \(x\) (its signed position from a fixed origin \(O\)) is a function of time \(t\). Its motion is described by two derivatives:

Velocity is the rate of change of displacement, \(v=\dfrac{dx}{dt}\). Its sign gives the direction of motion; its size \(|v|\) is the speed.

Acceleration is the rate of change of velocity, \(a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}\) — the second derivative of displacement.

The particle is momentarily at rest when \(v=0\); it is moving in the positive direction when \(v>0\) and the negative direction when \(v<0\). It is speeding up when \(v\) and \(a\) share a sign and slowing down when they have opposite signs.

Key idea. \(x \xrightarrow{\;d/dt\;} v \xrightarrow{\;d/dt\;} a\). Differentiate to go from position to velocity to acceleration. At rest \(\Rightarrow v=0\); direction \(\Rightarrow\) sign of \(v\); speeding up \(\Rightarrow v\,a>0\).

Displacement-time: the gradient is the velocity, so the peak \((1,4)\) (where \(v=0\)) is the maximum displacement

Velocity-time: \(v=0\) at \(t=1\) and \(t=3\) — the particle is momentarily at rest, and \(v<0\) between them

Velocity and acceleration from displacement \(x(t)\):

\[v=\dfrac{dx}{dt},\qquad a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}\]

v = \frac{d x}{d t}, a = \frac{d^{2} x}{d t^{2}}

Reading the motion from the signs:

\[v=0 \ \text{(at rest)},\quad v>0 \ \text{(positive direction)},\quad v<0 \ \text{(negative direction)},\quad \text{speed}=|v|\]

Speeding up or slowing down. Compare the signs of \(v\) and \(a\): if \(v\,a>0\) (same sign) the particle is speeding up; if \(v\,a<0\) (opposite signs) it is slowing down.

Turning points of the motion (for maxima/minima):

\[\text{max/min displacement: } v=\dfrac{dx}{dt}=0;\qquad \text{max/min velocity: } a=\dfrac{dv}{dt}=0\]

a = \frac{d v}{d t} = 0

How to analyse straight-line motion

Differentiate. From \(x(t)\), find \(v=\dfrac{dx}{dt}\) and \(a=\dfrac{dv}{dt}\). Evaluate at a required time by substitution.
At rest. Solve \(v=0\) for the time(s); substitute back into \(x\) or \(a\) for the position or acceleration then.
Direction. Read the sign of \(v\): positive means the positive direction, negative means the negative direction.
Speeding up or slowing down. Compare the signs of \(v\) and \(a\): same sign \(\Rightarrow\) speeding up; opposite signs \(\Rightarrow\) slowing down.
Maxima/minima. For greatest displacement solve \(v=0\); for greatest/least velocity solve \(a=0\). Confirm with a second-derivative or sign test and check any interval endpoints.

Distance vs displacement. If the particle changes direction (at a time when \(v=0\)), the distance travelled is the sum of the distances between successive positions, not just the change in displacement.

Example 1 — Velocity at an instant

A particle has displacement \(x=2t^{3}-3t^{2}+4\) metres. Find its velocity when \(t=2\) s.

Solution

Velocity — differentiate the displacement:

\(v=\dfrac{dx}{dt}\)

\(=\)

\(6t^{2}-6t\)

Evaluate — substitute \(t=2\):

\(v(2)\)	\(=\)	\(6(2)^{2}-6(2)\)
	\(=\)	\(24-12\)
	\(=\)	\(12\)

\(\therefore\) Velocity at \(t=2\): \(12\;\text{m/s}\)

v = 12

Example 2 — When is it at rest?

A particle has displacement \(x=t^{3}-6t^{2}+9t+1\) metres. Find when it is momentarily at rest.

Solution

Velocity — differentiate the displacement:

\(v=\dfrac{dx}{dt}\)

\(=\)

\(3t^{2}-12t+9\)

At rest — set \(v=0\) and solve:

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

\(\therefore\) Momentarily at rest when \(t=1\) s and \(t=3\) s

t = 1 or 3

Example 3 — Speeding up or slowing down

For \(x=t^{3}-3t^{2}-9t+2\) metres, is the particle speeding up or slowing down at \(t=2\) s?

Solution

Velocity and acceleration — differentiate once and twice:

\(v=\dfrac{dx}{dt}\)	\(=\)	\(3t^{2}-6t-9\)
\(a=\dfrac{dv}{dt}\)	\(=\)	\(6t-6\)

Evaluate at \(t=2\):

\(v(2)\)	\(=\)	\(3(2)^{2}-6(2)-9\)
	\(=\)	\(12-12-9=-9\)
\(a(2)\)	\(=\)	\(6(2)-6\)
	\(=\)	\(6\)

Compare signs — form the product \(v\,a\):

\(v\,a\)	\(=\)	\((-9)(6)\)
	\(=\)	\(-54\)

\(\therefore\) \(v\,a<0\) (opposite signs), so the particle is slowing down

v a < 0

Example 4 — Maximum displacement

For \(x=t^{3}-6t^{2}+9t\) metres on \(0\le t\le 3\), find the maximum displacement.

Solution

Stationary points — differentiate and set \(v=0\):

\(v=\dfrac{dx}{dt}\)	\(=\)	\(3t^{2}-12t+9\)
	\(=\)	\(3(t-1)(t-3)\)
\(v=0\)	\(\Rightarrow\)	\(t=1\) or \(t=3\)

Classify \(t=1\) — use the acceleration \(a=6t-12\):

\(a(1)\)	\(=\)	\(6(1)-12\)
	\(=\)	\(-6<0\) (maximum)

Displacement there — substitute \(t=1\) into \(x\):

\(x(1)\)	\(=\)	\((1)^{3}-6(1)^{2}+9(1)\)
	\(=\)	\(1-6+9\)
	\(=\)	\(4\)

Check the endpoints of \(0\le t\le 3\):

\(x(0)\)	\(=\)	\(0\)
\(x(3)\)	\(=\)	\(27-54+27=0\)

\(\therefore\) Maximum displacement: \(4\;\text{m}\) (at \(t=1\) s)

x_{\max} = 4

Common pitfalls

Do not confuse velocity with acceleration. Velocity is the first derivative of displacement and acceleration is the second. Substituting a time into \(x\) or into the wrong derivative is the most common slip.

Slowing down is not the same as negative acceleration. A particle with \(a<0\) can still be speeding up (if \(v<0\) too). Compare the signs of \(v\) and \(a\): \(v\,a>0\) is speeding up, \(v\,a<0\) is slowing down.

Distance is not just the change in displacement. When the particle reverses direction (where \(v=0\)), add the distances between successive rest positions rather than subtracting the end displacements.

Frequently asked questions

How do you find velocity and acceleration from displacement?

Differentiate: \(v=\dfrac{dx}{dt}\) (once) and \(a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}\) (twice), then substitute the required time.

When is a particle momentarily at rest?

When \(v=0\). Set the velocity to zero and solve for \(t\); the particle stops there before changing direction.

How do you tell which direction a particle is moving?

By the sign of \(v\): \(v>0\) is the positive direction, \(v<0\) the negative direction, \(v=0\) momentarily at rest.

What is the difference between speeding up and slowing down?

Speeding up: \(v\) and \(a\) have the same sign (\(v\,a>0\)). Slowing down: opposite signs (\(v\,a<0\)).

How do you find the maximum displacement or velocity?

Maximum displacement is where \(v=0\); maximum/minimum velocity is where \(a=0\). Confirm with a second-derivative test and check the interval endpoints.

What is the difference between distance and displacement?

Displacement is the signed change in position; distance is the total path length. When the particle reverses, add the distances between successive rest positions.

Motion in a straight line

📖 Theory

How to analyse straight-line motion

Common pitfalls

Frequently asked questions

Theory