e.g. A supercar will accelerate to 50km/hr faster than a cyclist. That is to say, the supercar has a greater acceleration.

Calculating Acceleration

\[\begin{aligned}
acceleration &= \frac{\text{change in speed}}{\text{change in time}} \\
a &= \frac{\Delta v}{\Delta t} \\
\end{aligned}\]

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 \(ms^{-2}\))

Rearranging Equations

\[\begin{aligned}
a &= \frac{\Delta v}{\Delta t} && \text{v is divided by t} \\
a \times \Delta t &= \Delta v && \text{Undo the divide by multiplying} \\
\Delta t &= \frac{\Delta v}{a} && \text{Undo the multiplication by dividing}
\end{aligned}\]\[\begin{aligned}
v &= \frac{\Delta d}{\Delta t} && \text{d is divided by t} \\
v \times \Delta t &= \Delta d && \text{Undo the divide by multiplying} \\
\Delta t &= \frac{\Delta d}{v} && \text{Undo the multiplication by dividing}
\end{aligned}\]

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?

\[\begin{aligned}
9.8 &= \frac{54 - 0}{t} \\
9.8 \times t &= 54 \\
t &= \frac{54}{9.8} = 5.51s
\end{aligned}\]

Question 4

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:\(v_{i} = 5.55ms^{-1}\), \(v_{f} = 6.3ms^{-1}\), \(\Delta t = 3s\)