Motion and Motion Graphs: Question 3

Worked solution

Part (a): Interpreting the sloping section

Between $t = 0$ and $t = 8\text{ s}$, the graph is a straight, sloping line rising from $0\text{ m/s}$ to $20\text{ m/s}$. Because it is a straight line, the speed is increasing at a constant (uniform) rate. This section shows uniform acceleration.

Part (b): Distance during the first section

The area between the graph and the time axis gives the distance travelled. For $t = 0$ to $t = 8\text{ s}$, this area is a triangle, with base $8\text{ s}$ and height $20\text{ m/s}$:

$\text{distance} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8\text{ s} \times 20\text{ m/s}$

$\text{distance} = 80\text{ m}$

Part (c): Distance during the second section

For $t = 8\text{ s}$ to $t = 20\text{ s}$, the speed is constant at $20\text{ m/s}$, so this section of the graph is a rectangle, with width equal to the time taken:

$\text{time for this section} = 20\text{ s} - 8\text{ s} = 12\text{ s}$

$\text{distance} = \text{base} \times \text{height} = 12\text{ s} \times 20\text{ m/s}$

$\text{distance} = 240\text{ m}$

Part (d): Total distance for the whole test

The third section, from $t = 20\text{ s}$ to $t = 24\text{ s}$, is another triangle, as the speed decreases uniformly from $20\text{ m/s}$ to $0\text{ m/s}$ over $4\text{ s}$:

$\text{distance}_3 = \frac{1}{2} \times 4\text{ s} \times 20\text{ m/s} = 40\text{ m}$

The total distance for the whole test is the sum of the areas of all three sections:

$\text{total distance} = 80\text{ m} + 240\text{ m} + 40\text{ m}$

$\text{total distance} = 360\text{ m}$

Final answers

• (a) The straight sloping line shows uniform (constant) acceleration
• (b) Distance in the first section $=$ $80\text{ m}$
• (c) Distance in the second section $=$ $240\text{ m}$
• (d) Total distance for the whole test $=$ $360\text{ m}$