Velocity and acceleration

Displacement, velocity and acceleration for motion in a straight line, distance versus displacement, average velocity and speed, when a particle is at rest, and when it is speeding up.

Displacement, velocity, acceleration

Quantity	Meaning	Units
Displacement $x$	Position relative to the origin	m
Velocity $v = \dfrac{dx}{dt}$	Rate of change of displacement. Its sign gives the direction.	m/s
Acceleration $a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$	Rate of change of velocity	m/s²

A worked example

x = t³ − 6t² + 9t

The horizontal axis is time t. Drag the point and watch v change sign at t = 1 and t = 3.

Example 1●●●

A particle moves with $x = t^3 - 6t^2 + 9t$ metres, for $t \ge 0$ seconds.

$v = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$ and $a = 6t - 12$.

Velocity and acceleration

$v = 0$ at $t = 1$ and $t = 3$, where $x = 4$ and $x = 0$.

At rest

It goes from $x = 0$ to $4$, back to $0$, then out to $x(4) = 4$. Distance $= 4 + 4 + 4 = 12$ m, but displacement is only $4$ m.

Distance in 4 seconds

Average velocity $= \dfrac{4}{4} = 1$ m/s. Average speed $= \dfrac{12}{4} = 3$ m/s.

Averages

For $1 < t < 2$, $v < 0$ and $a < 0$, so it speeds up. For $t > 3$, both are positive, so it speeds up again.

Speeding up

Flashcards

$v$ and $a$ from $x$

$v = \dfrac{dx}{dt}$, $\;a = \dfrac{d^2x}{dt^2}$

At rest

$v = 0$

Speeding up

$v$ and $a$ have the same sign.

Practice

For $x = t^3 - 6t^2 + 9t$, find the initial velocity.
When is that particle at rest?
Find its acceleration at $t = 3$.
When is its acceleration zero?
Find the distance travelled in the first $4$ seconds.
Find the displacement after $4$ seconds.
Find the average speed over the first $4$ seconds.
A particle is speeding up when:
The units of acceleration are: