Mastering Distance-Time and Velocity-Time Graphs for Your Physics Exam

What the examiner is testing

Examiners assess your ability to extract specific information from graphs, such as speed, acceleration, and total distance travelled. Marks are awarded for correctly calculating gradients and areas under the curve, demonstrating an understanding of how these mathematical operations relate to physical quantities.

The method

1. Identify the graph type: Determine if it's a distance-time graph (distance on y-axis, time on x-axis) or a velocity-time graph (velocity on y-axis, time on x-axis). This is crucial as the interpretation of gradient and area differs between them.
2. Recall what the gradient represents: For a distance-time graph, the gradient represents speed. For a velocity-time graph, the gradient represents acceleration.
3. Recall what the area under the graph represents: For a velocity-time graph, the area under the graph represents the total distance travelled. There is no physical meaning to the area under a distance-time graph.
4. Calculate the gradient (if required): Use the formula \( \text{gradient} = \frac{\text{change in y-axis quantity}}{\text{change in x-axis quantity}} \). Select two clear points on the line segment you are interested in.
5. Calculate the area under the graph (if required): Divide the area into standard geometric shapes (rectangles, triangles, trapezoids). Calculate the area of each shape using standard formulae and sum them to find the total area.
6. State units: Always include the correct units for your calculated quantity (e.g., m/s for speed, m/s\(^2\) for acceleration, m for distance).

Worked example

A car accelerates uniformly from rest to \(15 \text{ m/s}\) in \(5 \text{ s}\). Calculate the acceleration of the car.

$$ \text{Acceleration} = \frac{\text{change in velocity}}{\text{change in time}} $$
$$ \text{Acceleration} = \frac{15 \text{ m/s} - 0 \text{ m/s}}{5 \text{ s} - 0 \text{ s}} $$
$$ \text{Acceleration} = \frac{15 \text{ m/s}}{5 \text{ s}} $$
$$ \text{Acceleration} = 3 \text{ m/s}^2 $$
Sanity check: The car is speeding up, so a positive acceleration is expected. \(3 \text{ m/s}^2\) is a reasonable acceleration for a car.

Worked example: a harder one

A car travels at a constant speed of \(20 \text{ m/s}\) for \(10 \text{ s}\), then brakes uniformly to a stop in another \(5 \text{ s}\). Calculate the total distance travelled.

Initial thought: Just use distance = speed \(\times\) time for the whole journey.
Why this fails: The speed is not constant throughout the entire journey. This formula only works for constant speed.

We need to use a velocity-time graph approach, even if not explicitly drawn.
The journey consists of two parts:
Part 1: Constant velocity. This forms a rectangle on a velocity-time graph.
Part 2: Uniform deceleration. This forms a triangle on a velocity-time graph.

Distance for Part 1 (rectangle):
$$ \text{Area}_1 = \text{length} \times \text{width} $$
$$ \text{Area}_1 = \text{velocity} \times \text{time} $$
$$ \text{Area}_1 = 20 \text{ m/s} \times 10 \text{ s} $$
$$ \text{Area}_1 = 200 \text{ m} $$

Distance for Part 2 (triangle):
The initial velocity for this part is \(20 \text{ m/s}\), final velocity is \(0 \text{ m/s}\), and the time taken is \(5 \text{ s}\).
$$ \text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} $$
$$ \text{Area}_2 = \frac{1}{2} \times \text{time} \times \text{change in velocity} $$
$$ \text{Area}_2 = \frac{1}{2} \times 5 \text{ s} \times 20 \text{ m/s} $$
$$ \text{Area}_2 = 50 \text{ m} $$

Total distance travelled:
$$ \text{Total distance} = \text{Area}_1 + \text{Area}_2 $$
$$ \text{Total distance} = 200 \text{ m} + 50 \text{ m} $$
$$ \text{Total distance} = 250 \text{ m} $$
Sanity check: The car travels for a total of \(15 \text{ s}\). If it travelled at a constant \(20 \text{ m/s}\) for the whole time, it would cover \(300 \text{ m}\). Since it slowed down, \(250 \text{ m}\) is a reasonable, smaller value.

Practice

1. A car's journey is represented by the distance-time graph below.

   Distance (m) ^ | / | / | / | / +-----------> Time (s) 0 2 4 6
   If the car travels \(10 \text{ m}\) in \(5 \text{ s}\), what is its speed?

2. A train accelerates uniformly from \(10 \text{ m/s}\) to \(30 \text{ m/s}\) in \(10 \text{ s}\). Calculate its acceleration.

3. A cyclist travels at a constant velocity of \(8 \text{ m/s}\) for \(15 \text{ s}\). What is the total distance travelled?

4. The velocity-time graph below shows the motion of a scooter.

   Velocity (m/s) ^ | ___________ | / \ | / \ | / \ +-------------------> Time (s) 0 5 10 15 20 25
   The scooter accelerates from rest to \(10 \text{ m/s}\) in \(5 \text{ s}\), travels at a constant velocity for \(10 \text{ s}\), and then decelerates uniformly to rest in \(10 \text{ s}\). Calculate the total distance travelled by the scooter.

Answers:

1. \(2 \text{ m/s}\)
2. \(2 \text{ m/s}^2\)
3. \(120 \text{ m}\)
4. Working for question 4:
   The graph can be divided into three sections:
   Section 1 (0-5 s): Triangle
   $$ \text{Distance}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \text{ s} \times 10 \text{ m/s} = 25 \text{ m} $$
   Section 2 (5-15 s): Rectangle
   $$ \text{Distance}_2 = \text{base} \times \text{height} = (15 - 5) \text{ s} \times 10 \text{ m/s} = 10 \text{ s} \times 10 \text{ m/s} = 100 \text{ m} $$
   Section 3 (15-25 s): Triangle
   $$ \text{Distance}_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (25 - 15) \text{ s} \times 10 \text{ m/s} = \frac{1}{2} \times 10 \text{ s} \times 10 \text{ m/s} = 50 \text{ m} $$
   Total distance = \(\text{Distance}_1 + \text{Distance}_2 + \text{Distance}_3 = 25 \text{ m} + 100 \text{ m} + 50 \text{ m} = 175 \text{ m}\).

The three mistakes that lose marks

1. Confusing graph types: Using the gradient of a distance-time graph to find acceleration, or the area under a distance-time graph to find distance.
   Wrong answer example: For a distance-time graph showing a curved line, calculating the gradient at a point and calling it acceleration. (It's instantaneous speed).
2. Incorrectly calculating area under a velocity-time graph: Misidentifying shapes or using the wrong formula for an area.
   Wrong answer example: For a trapezoidal section of a velocity-time graph, calculating the area as a rectangle using the average velocity and total time, but forgetting to account for the initial or final velocity properly, or simply using length \(\times\) width with the peak velocity.
3. Forgetting units or using incorrect units: Not including units with your final answer, or using units that don't match the quantity calculated.
   Wrong answer example: Stating an acceleration of "5", or "5 m/s" instead of "5 m/s\(^2\)".

30-second recap

Distance-time graphs show how far an object has travelled over time; their gradient is speed. Velocity-time graphs show an object's velocity over time; their gradient is acceleration, and the area under them is the total distance travelled. Always identify the graph type first and apply the correct interpretation for gradient and area.