Average velocity and average speed

Displacement as a function of time, graphing motion, average velocity, distance travelled, and average speed.

Displacement and time

A particle moves along a number line. Its displacement $x$ is its position relative to the origin, and it can be negative. The distance travelled counts every metre moved, in either direction.

Average velocity and speed

$$\text{average velocity} = \frac{\text{change in displacement}}{\text{change in time}} \qquad \text{average speed} = \frac{\text{distance travelled}}{\text{time taken}}$$

Example 1●●

A particle has displacement $x = t^2 - 6t + 5$ metres after $t$ seconds. Find, for $0 \le t \le 5$, the average velocity, the distance travelled and the average speed.

$x(0) = 5$, $x(3) = -4$, $x(5) = 0$.

Average velocity $= \dfrac{0 - 5}{5} = -1$ m/s.

Distance $= 9 + 4 = 13$ m.

It moves left from 5 to −4 until t = 3, then turns and moves right to 0.

Average speed $= \dfrac{13}{5} = 2.6$ m/s.

Displacement–time graph

Average velocity is the gradient of the dashed chord from t = 0. Move P along the curve.

Try this

At what time is the average velocity since $t = 0$ equal to $-3$ m/s?

Flashcards

Average velocity

$\dfrac{\Delta x}{\Delta t}$ (can be negative)

Average speed

Distance $\div$ time (never negative)

Distance travelled

Add the lengths of each leg of the journey.

Practice

For $x = t^2 - 6t + 5$, find $x$ when $t = 1$.
For the same particle, find the average velocity for $0 \le t \le 5$.
Find the distance it travels for $0 \le t \le 5$.
Find its average speed for $0 \le t \le 5$.
Which can be negative?
Find the average velocity during the first second, $0 \le t \le 1$.
A ball has height $h = 20t - 5t^2$ metres. Find its average velocity from $t = 1$ to $t = 3$.
Find the distance the ball travels from $t = 1$ to $t = 3$.
Displacement measures: