Motion Review and Basic Vectors

Kinematics in 1D – Review

This unit builds on the skills from the first and fifth units in SPH3U and uses the same formulae. Below are some typical questions for one-dimensional motion to help you remember how to interpret questions and use the formulae. For a more comprehensive review, see here.

Examples

Example 1

Determine the acceleration of a car that slows from 75 km/h to a stop in 9.5 seconds.

Solution

    \[\vec{v}_1 = 75 km/h\]

    \[\vev{v}_2 = 0\]

    \[\Delta t = 9.5 s\]

    \[\vec{a} = ?\]

 

Don’t forget to convert to the right units!

    \[\begin{align*} \vec{v}_1 &= 75 km/h \times \frac{1000 m}{1 km} \times \frac{1 h}{3600 s} \\ &\approx 20.8 m/s\end{align*}\]

 

It’s now time to calculate.

    \[\begin{align*} \vec{a} &= \frac{\vec{v}_2 - \vec{v}_1}{\Delta t} \\ &= \frac{0 - 20.8}{9.5} \\ &\approx -2.2 m/s^2\end{align*}\]

 

Finally, remember the decimal rule for reporting final answers.

 

Example 2

Determine how far the car in question 1 travels as it slows to a stop.

Solution

    \[\begin{align*} \Delta\vec{d} &= \vec{v}_2 \Delta t - \frac{1}{2}\vec{a}\Delta t^2 \\ &= 0(9.5) - \frac{1}{2}(-2.2)(9.5)^2 \\ &\approx 99.3 m\end{align*}\]

 

Example 3

A cyclist accelerates from 5.0 m/s at 1.5 m/s2. Determine how fast the cyclist is travelling in meters per second and kilometers per hour after accelerating for 22.0 m.

Solution

    \[\vec{v}_1 = 5.0 m/s\]

    \[\vec{a} = 1.5 m/s^2\]

    \[\Delta\vec{d} = 22.0 m\]

    \[\vec{v}_2 = ?\]

 

    \[\begin{align*} (\vec{v}_2)^2 &= (\vec{v}_1)^2 + 2\vec{a}\Delta\vec{d} \\ &= 5.0^2 + 2(1.5)(22.0) \\ \vec{v}_2 &= \sqrt{91} \\ &\approx 9.5 m/s\end{align*}\]

 

    \[\begin{align*} \vec{v}_2 &= 9.5 m/s \times \frac{1 km}{1000 m} \times \frac{3600 s}{1 h} \\ &\approx 34.2 km/h\end{align*}\]

 

Example 4

Sprinting to class, Peter travels 18.0 m in 3.0 seconds. What is Peter’s average velocity?

Solution

    \[\Delta\vec{d} = 18.0 m\]

    \[\Delta t = 3.0 s\]

    \[\vec{v}_{avg} = ?\]

 

    \[\begin{align*} \vec{v}_{avg} &= \frac{\Delta\vec{d}}{\Delta t} \\ &= \frac{18.0 m}{3.0 s} \\ &= 6.0 m/s\end{align*}\]

Vector Addition and Subtraction in 1D

Vector addition and subtraction is a key part to calculating motion in two dimensions. For one-dimensional vectors, it might be helpful to review what negative numbers mean in physics.

Example

MJ is confused. First she heads 7.5 m [N] to head to the kitchen and get some cookies. Then, she gets distracted by American Idol on TB and she moves back towards the living room (3.0 m [S]). During the commercials, she finished the trip to the kitchen, travelling a further 6.0 m [N]. Determine her overall displacement.

Solution

Finding an overall vector is as simple as adding up all of the pieces that make it up. (This also works the other way, you can split a vector into separate pieces, but more on that later). First, you must define a positive direction:

    \[[N]+\]

 
List each vector and figure out whether it is positive or negative. In the future, we will draw the vectors as well.

    \[\Delta\vec{d}_1 = 7.5 m\]

    \[\Delta\vec{d}_2 = -3.0 m\]

    \[\Delta\vec{d}_3 = 6.0 m\]

 

Then, as long as all your measurements are in the same unit, you add:

    \[\begin{align*} \Delta\vec{d} &= \Delta\vec{d}_1 + \Delta\vec{d}_2 + \Delta\vec{d}_3 \\ &= 7.5 + (-3.0) + 6.0 \\ &= 10.5 m [N]\end{align*}\]

Additional Resources

Assignment Questions

Assignment 1:

  • Pg. 20, #2, 6
  • Pg. 27, #20, 22-24