2.16.2 Speed-Time Graph
A speed-time graph, also known as a velocity-time graph, is a graphical representation of the relationship between an object’s speed and time during its motion.
- The graph shows the change in an object’s speed over time, with speed on the vertical axis (\(y-axis\)) and time on the horizontal axis (\(x-axis\)).
- A straight horizontal line represents constant speed, while a curved line represents changing speed.
- The graph shows the change in an object’s speed over time, with speed on the vertical axis (\(y-axis\)) and time on the horizontal axis (\(x-axis\)).
- The slope of the line represents acceleration or deceleration, with a steeper slope indicating a greater acceleration or deceleration.
- The area under the line in a speed-time graph represents the distance traveled by the object during its motion.
- To calculate the distance traveled
- Find the area of the trapezoid under the line by taking the average speed and multiplying it by the time traveled.
- To calculate the distance traveled
- In addition, the slope of the line in a speed-time graph can be used to calculate the acceleration of the object, using the formula
\(acceleration =\frac{change\ in\ speed}{time}\)
- If the slope is positive, it represents acceleration, while a negative slope represents deceleration.
- If the slope is zero, the object is moving at a constant speed.
Worked example:
The diagram shows speed-time graph for the first 40 seconds of a cycle ride.
Find the acceleration between 20 and 40 seconds.
Acceleration is equal to the gradient of the line between 20 and 40 seconds.
Find the gradient by using
\(m=\frac{rise}{run}\)
\(m=\frac{4-2}{40-20}\)
\(m=\frac{2}{20}\)
\(m=0.1\)
\(Acceleration=0.1 \frac{m}{s^2}\)
Tip:
It is simple to become confused between various graph types.
- Check the label on the vertical axis to ensure that you’re looking at a speed-time graph or a distance-time graph
Worked example:
The diagram shows the speed-time graph for 70 seconds of a car journey.
Calculate the deceleration of the car during the first 20 seconds.
Deceleration is equal to the gradient of the line between 0 and 20 seconds.
Find the gradient by using
\(m=\frac{(y^2-y^1)}{(x^2-x^1)}\)
\(m=\frac{10-16}{20-0}\)
\(m=-\frac{6}{20}\)
\(m=-0.3\)
\(-0.3\frac{m}{s^2}\)
Calculate the total distance traveled by the car during the 70 seconds.
The area under the line in a speed-time graph represents the distance.
Let’s divide the area in parts
\(A1\) is the area of a right triangle and \(A2\) is the area of a rectangle.
\(-0.3\frac{m}{s^2}\)
\(A_1=\frac{1}{2}(h×b)\)
\(A_1=\frac{1}{2}((16-10)×20)\)
\(=\frac{1}{2}(6×20)\)
\(=\frac{1}{2}(120)\)
\(A_1=60\)
\(A_2=l×w\)
\(A_2=70×10\)
\(A_2=700\)
The total area under the curve is
\(A_1+A_2=60+700\)
\(=760\)
So, the required distance is \(760m\)
Test yourself
Question 1:
The diagram shows information about the final 70 seconds of a car journey.
- Find the acceleration of the car between 60 and 70 seconds.
[1]
- Find the distance traveled by the car during the 70 seconds.
[3]
Question 2:
The diagram shows the speed-time graph for the final 40 seconds of a car journey.
At the start of the 40 seconds the speed is v m/s .
The total distance traveled during the 40 seconds is 1.24 kilometers.
Find the value of v.
[4]
Question 3:
The diagram shows the speed-time graph for 100 seconds of the journey of a
car and of a motorbike.
Calculate how much further the car traveled than the motorbike during the 100 seconds.
[3]
Question 4:
The diagram shows the speed-time graph of a train journey between two stations.
- Find the acceleration of the train during the first 40 seconds.
[1]
- Calculate the distance between the two stations.
[3]
Question 5:
The diagram shows the speed-time graph for part of a journey for two people, a runner and a walker.
Calculate the acceleration of the runner for the first 3 seconds.
[1]
Calculate the total distance traveled by the runner in the 19 seconds.
[3]
The runner and the walker are traveling in the same direction along the same path.
When \(t=0\), the runner is 10 meters behind the walker.
Find how far the runner is ahead of the walker when \(t=19\).