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When you do nothing, you feel overwhelmed and powerless. But when you get involved, you feel the sense of hope and accomplishment that comes from knowing you are working to make things better.
Pauline R. Kezer
Consider the photograph taken of Sven riding his bicycle down a hill. What do you think the position-time graph of his motion will look like? He is moving forwards, so we can assume that the graph will go up and to the right, but how steep will it be? At the top of the hill he is just getting started and will not be moving very quickly, so the graph should not have a very steep slope. However, as he rides farther down the hill he will be moving faster. We show faster velocity by drawing a graph with a steeper slope.The position-time graph of Sven’s ride down the hill is shown below.
Sven’s Ride Down the Hill
If you look closely at the position-time graph on the left, you can see that the slope of the black line gradually gets steeper as time increases. At t = 1 s, the slope of the line is very small (the green line) and he is not moving very quickly. At t = 5 s, the slope of the line is quite steep (the red line) and he is moving quite fast. We could find Sven’s instantaneous velocity by calculating the slope of the tangent line at that particular point on the graph.
Whenever the velocity of an object changes, we say that it is accelerating.
Velocity-Time Graphs
A velocity-time (v-t) graph can be used to show how Sven’s velocity changes over time. A v-t graph of Sven’s ride down the hill is shown below.
Acceleration is equal to the slope of a velocity-time graph.
By choosing two points on the v-t graph, Sven’s acceleration down the hill can be determined.
Sven accelerates down the hill at 2 m/s2.
Practice Questions
- Determine the acceleration for each of the three cases shown to the right.How would you draw in a line that shows an acceleration of 1.5 m/s2 starting at the origin (0, 0)?
Answer
Blue line:
The acceleration is 0.67 m/s2. This means that the velocity is becoming more positive and the object is speeding up.
Green line:
The acceleration is 0 m/s2. This means that the velocity does not change. The object may be stationary or moving at a constant speed.
Purple line:
The acceleration is -0.5 m/s2. This means that the velocity is becoming more negative and the object is slowing down.
The orange line in the image below shows an acceleration of 1.5 m/s2 starting at the origin (0, 0). Notice that at 1s, v=1.5 m/s2; at 2s, v= 3.0 m/s2; and 3s, v=4.5 m/s2. In other words, the object’s speed increases by 1.5 m/s every second.
Finding displacement from a v-t graph
This graph shows the motion of an object travelling at 3 m/s for 9 s. Since the velocity is constant, the displacement can be found by using the equation:
This value can also be found by finding the area under the v-t graph as shown below.
Example: Finding the Displacement of an Accelerating Object
The graph to the right shows the motion of a car accelerating from rest on a country road.A) Calculate the acceleration of the car.B) Find the car’s displacement over the first 8 s.
- Acceleration is equal to the slope of the v-t graph. If the two end-points are chosen, (0, 0) and (8, 20), then the acceleration can be found as follows:
The acceleration of the car is 2.5 m/s2.
- Displacement can be found from a v-t graph by determining that area under the line of the graph.
The car’s displacement over the 8 s interval is 80 m [N].
Practice Questions
- Find the displacement of the object whose motion is shown in the v-t graph on the left.
- Describe the motion of the object.
Answer
A)
By dividing the graph up into triangles and rectangles, it will be easier to calculate the area under the graph.
A)
Area 1 = ½(base)(height) = ½(4 s)(4 m/s) = 8 m
Area 2 = (base)(height) = (2 s)(4 m/s) = 8 m
Area 3 = ½(base)(height) = ½(2 s)(4 m/s) = 4 m
Area 4 = ½(base)(height) = ½(1 s)(-2 m/s) = -1 m. All values are read off of the axis. Therefore, in this case, the negative velocity (the object is moving backwards) results in a negative displacement.
Area 5 = (base)(height) = (3 s)(-2 m/s) = -6 m
Displacement = total area = Area 1 + Area 2 + Area 3 + Area 4 + Area 5
Displacement = 8 m + 8 m + 4 m + (-1 m) + (-6 m)
Displacement = 13 m
B)
The object undergoes an acceleration of 1 m/s2 for 4 seconds, travels at 4 m/s for 2 s, accelerates at -2 m/s2 (really a deceleration of 2m/s2) for 3 s, and then continues at -2 m/s for 3 s.
The object is moving forwards between t = 0 s and t = 8 s. It is moving backwards from a time of t = 8 s to t = 12 s. At t =8 s, the object is not moving (v = 0 m/s).