Example
What is the average speed and average velocity for this trip?
The trip covers a total of 40 km in 70 minutes.
\[
\text{Average speed}
= \frac{\text{total distance}}{\text{total time}}
= \frac{40}{70/60}
= 34.3\text{ km/h}
\]
Average velocity is the rate of change of displacement with time.
Velocity is the gradient of a displacement-time graph.
\[
v_{\text{avg}}
= \frac{\text{displacement}}{\text{time interval}}
= \frac{x_2 - x_1}{t_2 - t_1}
= \frac{\Delta x}{\Delta t}
\]
\[
\text{Average velocity}
= \frac{\Delta s}{\Delta t}
\]
The total displacement is zero, so the average velocity is \(0\text{ km/h}\).
15 km were covered in the first 30 minutes:
\[
\text{Speed} = \frac{15}{0.5} = 30\text{ km/h}
\]
A 10‑minute rest took place.
Then the furthest point was reached after another 10 minutes:
\[
\text{Speed} = \frac{10}{\tfrac{1}{6}} = 60\text{ km/h}
\]
Finally, 15 km were covered in 10 minutes:
\[
\text{Speed} = \frac{15}{\tfrac{1}{6}} = 90\text{ km/h}
\]
Instantaneous Speed and Velocity
Instantaneous velocity is the derivative of displacement:
\[
v(t) = \frac{ds}{dt}
\]
Instantaneous speed is the magnitude of velocity:
\[
\text{speed} = |v(t)|
\]
Velocity Function for the Journey
The velocity is piecewise constant:
\[
v(t)=
\begin{cases}
30 & 0 \le t \lt 30 \\
0 & 30 \le t \lt 40 \\
60 & 40 \le t \lt 50 \\
-90 & 50 \le t \le 60
\end{cases}
\]
Area Under a Velocity–Time Graph
The area under the graph gives displacement:
\[
s = \int v(t)\,dt
\]
Example
\[
\text{Area} = \tfrac12 bh
\]
\[
\text{Area}
= lb
= 0.5 \times 30
= 15\ \text{km}
\]
\[
s = \int v\,dt
\]
\[
= \int_{0}^{0.5} 30\,dt
\]
\[
= \left[\,30t\,\right]_{0}^{0.5}
\]
\[
= (30 \times 0.5) - (30 \times 0)
\]
\[
= 15\ \text{km}
\]
\[
\text{Area}
= lb
= \left(\frac{50}{60} - \frac{40}{60}\right)\times 30
\]
\[
= \frac{1}{6}\times 30
= 5\left(\frac{\text{km}}{\text{h}} \times \text{h}\right)
= 5\ \text{km}
\]
\[
s = \int v\,dt
\]
\[
= \int_{\tfrac{2}{3}}^{\tfrac{5}{6}} 30\,dt
\]
\[
= \left[\,30t\,\right]_{\tfrac{2}{3}}^{\tfrac{5}{6}}
\]
\[
= \left(30 \times \frac{5}{6}\right)
- \left(30 \times \frac{2}{3}\right)
\]
\[
= 25 - 20
= 5\ \text{km}
\]
\[
\text{A negative area is meaningless, so we take the modulus.}
\]
\[
\text{Area} = |lb|
\]
\[
= \left|\left(\frac{60}{60} - \frac{50}{60}\right)\times(-30)\right|
\]
\[
= \left|\frac{1}{6}\times(-30)\right|
\]
\[
= \left|-5\left(\frac{\text{km}}{\text{h}}\times\text{h}\right)\right|
\]
\[
= |-5\text{ km}|
= 5\text{ km}
\]
\[
s = \int v\,dt
\]
\[
= \int_{\tfrac{5}{6}}^{1} (-30)\,dt
\]
\[
= \left[\, -30t \,\right]_{\tfrac{5}{6}}^{1}
\]
\[
= (-30) - \left(-30 \times \frac{5}{6}\right)
\]
\[
= -30 + 25
= -5\ \text{km}
\]
\[
\text{Area}
= |lb|
\]
\[
= \left|\left(\frac{70}{60} - \frac{60}{60}\right)\times(-90)\right|
\]
\[
= \left|\frac{1}{6}\times(-90)\right|
\]
\[
= 15\ \text{km}
\]
\[
s = \int v\,dt
\]
\[
= \int_{1}^{\tfrac{7}{6}} (-90)\,dt
\]
\[
= \left[\, -90t \,\right]_{1}^{\tfrac{7}{6}}
\]
\[
= \left(-90 \times \frac{7}{6}\right)
- \left(-90 \times 1\right)
\]
\[
= -105 + 90
= -15\ \text{km}
\]
Acceleration is the gradient of a velocity-time graph.
Average acceleration is the rate of change of velocity with time.
\[
a_{\text{avg}} = \frac{\Delta v}{\Delta t}
\]
The instantaneous acceleration is the rate of change of velocity at that particular time.
\[
a(t) = \frac{dv}{dt}
\]
Example
The velocity-time graph of an item is plotted.
The item starts from rest.
After 8 seconds, the item is travelling at a velocity of 120m/s, which is maintained for a further 7 seconds.
After 30 seconds, the item is travelling at -75m/s.
Find the acceleration for each part of the journey.
From the graph:
\[
a = \frac{dv}{dt}
\]
\[
= \frac{120 - 0}{8}
\]
\[
= 15\,\frac{m/s}{s}
\]
\[
= 15\,\frac{m}{s^{2}}
\quad\text{or}\quad
15\,m\,s^{-2}
\]
\[
a = \frac{dv}{dt}
\]
\[
= \frac{120 - 120}{7}
\]
\[
= 0\,m\,s^{-2}
\]
\[
a = \frac{dv}{dt}
\]
\[
= \frac{-75 - 120}{15}
\]
\[
= \frac{-195}{15}
\]
\[
= -13\,m\,s^{-2}
\]
Deriving the Equations of Motion
For uniform acceleration:
\[
a_{\text{avg}} = \text{Instantaneous acceleration}
\]
\[
\text{Let } v = \text{final velocity},
u = \text{initial velocity at } t = 0,
t = t_2 = \text{final time}.
\]
\[
a = \frac{\Delta v}{\Delta t}
\]
\[
= \frac{v_2 - v_1}{t_2 - t_1}
\]
\[
= \frac{v - u}{t - 0}
\]
\[
= \frac{v - u}{t}
\]
So
\[
a = \frac{v - u}{t}
\]
\[
at = v - u
\]
\[
at + u = v
\]
\[
v = u + at
\]
For uniform acceleration, the average velocity is the mean of the intitial and final velocity:
\[
v_{\text{avg}} = \frac{v + u}{2}
\]
Velocity is the rate of change of displacement with time.
\[
v_{\text{avg}} = \frac{\Delta x}{\Delta t}
\]
\[
= \frac{x_2 - x_1}{t_2 - t_1}
\]
\[
\text{when } t = 0
\]
\[
v_{\text{avg}} = \frac{x_2 - x_1}{t}
\]
\[
v_{\text{avg}}\, t + x_1 = x_2
\]
But
\[
v = u + at
\]
\[
v_{\text{avg}} = \frac{v + u}{2}
\]
\[
v_{\text{avg}} = \frac{(u + at) + u}{2}
\]
\[
v_{\text{avg}} = \frac{2u + at}{2}
\]
\[
v_{\text{avg}} = u + \frac{at}{2}
\]
so
\[
\left(u + \frac{at}{2}\right)t + x_1 = x_2
\]
\[
\left(u + \frac{at}{2}\right)t = x_2 - x_1
\]
\[
\text{but } s = x_2 - x_1
\]
\[
\text{so}
\]
\[
\left(u + \frac{at}{2}\right)t = s
\]
\[
s = ut + \frac{at^2}{2}
\]
\[
v = u + at
\]
\[
t = \frac{v - u}{a}
\]
\[
s = ut + \frac{at^2}{2}
\]
\[
s = u\left(\frac{v-u}{a}\right)
+ \frac{a}{2}\left(\frac{v-u}{a}\right)^2
\]
\[
s = \frac{uv - u^2}{a}
+ \frac{a}{2}
\left(\frac{v^2 - 2uv + u^2}{a^2}\right)
\]
\[
s = \frac{uv - u^2}{a}
+ \frac{v^2 - 2uv + u^2}{2a}
\]
\[
= \frac{1}{2a}
\left[
2(uv - u^2)
+ v^2 - 2uv + u^2
\right]
\]
\[
= \frac{1}{2a}
\left[
-u^2 + v^2
\right]
\]
\[
= \frac{1}{2a}
\left(
v^2 - u^2
\right)
\]
\[
2as = v^2 - u^2
\]
\[
v^2 = u^2 + 2as
\]
Positive direction is upward.
Acceleration due to gravity:
\[
a = -g
\]
Magnitude:
\[
g = 9.8\text{ m/s}^2
\]
Do not substitute \(-9.8\) for \(g\); the sign is handled by direction.
© Alexander Forrest