Applications of the Motion Equations
What we have learnt in this past few lessons can be directly applied to road safety. We can analyse the relationship between speed and stopping distance. The following worked example illustrates this application. A truck is travelling at a constant velocity of 10 m/s ...
Equations of Motion: Applications in the Real-World
What we have learnt in this past few lessons can be directly applied to road safety. We can analyse the relationship between speed and stopping distance. The following worked example illustrates this application.
Example: Stopping Distance
Question
A truck is travelling at a constant velocity of \(\text{10}\) \(\text{m·s$^{-1}$}\) when the driver sees a child \(\text{50}\) \(\text{m}\) in front of him in the road. He hits the brakes to stop the truck. The truck accelerates at a rate of \(-\text{1.25}\) \(\text{m·s$^{-2}$}\). His reaction time to hit the brakes is \(\text{0.5}\) seconds. Will the truck hit the child?
Step 1: Analyse the problem and identify what information is given
It is useful to draw a time-line like this one:
We need to know the following:
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What distance the driver covers before hitting the brakes.
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How long it takes the truck to stop after hitting the brakes.
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What total distance the truck covers to stop.
Step 2: Calculate the distance \(AB\)
Before the driver hits the brakes, the truck is travelling at constant velocity. There is no acceleration and therefore the equations of motion are not used. To find the distance travelled, we use:
\begin{align*} v & = \frac{D}{t} \\ 10 & = \frac{D}{\text{0,5}} \\ D & = \text{5}\text{ m} \end{align*}The truck covers \(\text{5}\) \(\text{m}\) before the driver hits the brakes.
Step 3: Calculate the time \(BC\)
We have the following for the motion between \(B\) and \(C\):
\begin{align*} {\vec{v}}_{i} & = \text{10}\text{ m·s$^{-1}$} \\ {\vec{v}}_{f} & = \text{0}\text{ m·s$^{-1}$} \\ a & = -\text{1,25}\text{ m·s$^{-2}$} \\ t & = ? \end{align*}We can use Equation 1:
\begin{align*} {\vec{v}}_{f} & = {\vec{v}}_{i} + at \\ 0 & = \text{10}\text{ m·s$^{-1}$} + \left(-\text{1,25}\text{ m·s$^{-2}$}\right)t \\ -\text{10}\text{ m·s$^{-1}$} & = \left(-\text{1,25}\text{ m·s$^{-2}$}\right)t \\ t & = \text{8}\text{ s} \end{align*}Step 4: Calculate the distance \(BC\)
For the distance we can use Equation 2 or Equation 3. We will use Equation 2:
\begin{align*} \Delta \vec{x} & = \frac{\left({\vec{v}}_{i} + {\vec{v}}_{f}\right)}{2}t \\ \Delta \vec{x} & = \frac{10 + 0}{2} \left(8\right) \\ \Delta \vec{x} & = \text{40}\text{ m} \end{align*}Step 5: Write the final answer
The total distance that the truck covers is \({d}_{AB} + {d}_{BC} = \text{5} + \text{40} = \text{45}\) meters. The child is \(\text{50}\) meters ahead. The truck will not hit the child.
This lesson is part of:
One-Dimensional Motion