e.g. A supercar will reach 50km/hr faster than a
cyclist. That is to say, the supercar has a greater acceleration.
Calculating Acceleration
\[
\begin{aligned}
acceleration &= \frac{\text{change in speed}}{\text{change in
time}} \cr
a &= \frac{\Delta v}{\Delta t} \cr
\end{aligned}
\]
Velocity (\(v\)) has units meters per second (\(ms^{-1}\))
Time (\(t\)) has units seconds (\(s\))
Acceleration (\(a\)) has units meters per second per second
(\(m/s^{2}\) or \(ms^{-2}\))
Example
\(a = 2.5ms^{-2}\). A cyclist
accelerates from rest. They will gain \(2.5ms^{-1}\) of velocity every second!
t (s)
0
1
2
3
4
5
v (m/s)
0
2.5
5
7.5
10
12.5
Rearranging Equations
Work with a partner to rearrange \(a=\frac{\Delta v}{\Delta t}\) in your book
where \(v\) and \(t\) are the subject of the equation.
Whakatika
\[
\begin{aligned}
a &= \frac{\Delta v}{\Delta t} && \text{d is divided by
t} \cr
v \times \Delta t &= \Delta d && \text{Multiply both
sides by t} \cr
\end{aligned}
\]
\[
\begin{aligned}
a &= \frac{\Delta v}{\Delta t} && \text{v is divided by
t} \cr
a \times \Delta t &= \Delta v && \text{Multiply both
sides by t} \cr
\Delta t &= \frac{\Delta v}{a} && \text{Divide both
sides by a}
\end{aligned}
\]
Question 1
A bee starts at rest and takes off from a flower and reaches a
velocity of \(0.75ms^{-1}\) in \(0.5s\). Calculate it’s acceleration.
A skydiver at rest jumps out of a plane. They accelerate at \(9.8ms^{-2}\) until they reach a terminal
velocity of \(54ms^{-2}\). How long
does it take them to reach this speed?
A runner is approaching the finish line, moving at \(5.55ms^{-1}\) but needs to sprint to pass
the person just in front of them to get 1st place. They accelerate for
\(3s\) to reach \(6.3ms^{-1}\). What is their
acceleration?
Question 4: Answer
Knowns:\(v_{i} = 5.55ms^{-1}\), \(v_{f} = 6.3ms^{-1}\), \(\Delta t = 3s\)
