A distance-time graph, is a graphical representation of an object’s motion over a period of time. The graph plots the distance of the object along the vertical (y) axis and time along the horizontal (x) axis.
The slope of the graph represents the object’s velocity or speed. A steeper slope indicates a faster speed, while a flatter slope indicates a slower speed. If the slope of the graph is negative, it indicates that the object is moving in the opposite direction.
The area under the graph represents the total distance travelled by the object during the time interval represented by the graph.
Examples:
- The graph below shows the motion of a car.
a. What is the average speed of the car during the first 8 seconds?
b. What is the average speed of the car between 8 and 16 seconds?
c. What is the average speed of the car during the last 4 seconds?
d. What is the total distance traveled by the car?
2. The graph below shows the motion of a motorcyclist.
a. What is the total distance traveled by the motorcyclist?
b. What is the average speed of the motorcyclist during the first 6 seconds?
c. What is the average speed of the motorcyclist between 6 and 8 seconds?
d. What is the average speed of the motorcyclist between 10 and 14 seconds?
3. Plot the graph of the motion of a train that is travelling from point A to point B. The train starts at point A at 8:00 am and arrives at point B at 10:00 am, covering a distance of 120 kilometres. At point B the train stooped for 30 minutes and then returned with the same speed as before to point A. Halfway, the train stops for 10 minutes to let the passengers out. At what time will the train arrive to point A if it took 2.5 hours to return from point A to B? Assume that the train is moving at a constant speed throughout its journey.
a. What is the total distance traveled by the train?
b. What is the average speed of the train during the first 2 hours?
c. At what time the train will return back to point A?
Answers:
-
a. S= (D1-D2)/T S= (60-0)/8= 7.5 m/s
b. S= (D1-D2)/T S= (60-60)/8= 0 m/s the car is not moving, since the distance hasn’t changed.
c. S= (D1-D2)/T S= (60-0)/4= 15 m/s
d. Total distance = 60 m +60 m = 120 m
2.
a. Total distance = 400 m +400 m = 800 m
b. S= (D1-D2)/T S= (200-0)/6=33.33 m/s
c. S= (D1-D2)/T S= (400-200)/2=100 m/s
d. S= (D1-D2)/T S= (248-248)/4= 0 m/s the car is not moving, since the distance hasn’t changed.
3. Instructions:
- Draw a horizontal axis representing time and a vertical axis representing distance.
2. Mark the starting point of the train (point A) at the left end of the horizontal axis and the endpoint of the train (point B) at the right end of the horizontal axis. - Determine the time interval between the start and end of the journey and divide the horizontal axis into equal time intervals accordingly (e.g. if the journey is 2 hours, divide the horizontal axis into 2-hour intervals).
- Determine the scale for the vertical axis by considering the distance travelled by the train. For instance, if the train travels 120 kilometers, you may set the scale to 1 cm represents 10 kilometers.
- Plot the points for the train’s motion at equal time intervals. For example, if the journey is 2 hours, you may plot the points at 8:00 am, 9:00 am, and 10:00 am.
- Connect the points with a straight line to show the train’s motion over time.
a. Total distance = 120 km +120 km = 240 km
b. S= (D1-D2)/T S= (120-0)/2=60 km/h
c. 12:40 PM
Handouts: