Motion at Constant Velocity, Grade 10 Physics
Motion at a constant velocity or uniform motion means that the position of the object is changing at the same rate.

Assume that Lesedi takes 100 s to walk the 100 m to the taxi-stop every morning. If we assume that Lesedi’s house is the origin, then Lesedi’s velocity is:

Lesedi’s velocity is 1 m·s$^{−1}$. This means that he walked 1 m in the first second, another metre in the second second, and another in the third second, and so on. For example, after 50 s he will be 50 m from home. His position increases by 1 m every 1 s. A diagram of Lesedi’s position is shown in Figure 3.6. We can now draw graphs of position vs.time (x vs. t), velocity vs.time (v vs. t) and acceleration vs.time (a vs. t) for Lesedi moving at a constant velocity. The graphs are shown in Figure 3.7.

In the evening Lesedi walks 100 m from the bus stop to his house in 100 s. Assume that Lesedi’s house is the origin. The following graphs can be drawn to describe the motion.

Figure 3.8: Graphs for motion with a constant negative velocity (a) position vs. time (b) velocity vs. time (c) acceleration vs. time. The area of the shaded portion in the v vs.t graph corresponds to the object’s displacement.

We see that the v vs. t graph is a horisontal line. If the velocity vs. time graph is a horisontal line, it means that the velocity is constant (not changing). Motion at a constant velocity is known as uniform motion. We can use the x vs. t to calculate the velocity by finding the gradient of the line.

Lesedi has a velocity of -1 m·s$^{−1}$, or 1 m·s$^{−1}$ towards his house. You will notice that the v vs. t graph is a horisontal line corresponding to a velocity of -1 m·s$^{−1}$. The horisontal line means that the velocity stays the same (remains constant) during the motion. This is uniform velocity.
We can use the v vs. t to calculate the acceleration by finding the gradient of the line.

Lesedi has an acceleration of 0 m·s$^{−2}$. You will notice that the graph of a vs.t is a horisontal line corresponding to an acceleration value of 0 m·s$^{−2}$. There is no acceleration during the motion because his velocity does not change.

We can use the v vs. t to calculate the displacement by finding the area under the graph.

This means that Lesedi has a displacement of 100 m towards his house.
Exercise: Velocity and acceleration
E -1. Use the graphs in Figure 3.7 to calculate each of the following:
(a) Calculate Lesedi’s velocity between 50 s and 100 s using the x vs. t graph.
Hint: Find the gradient of the line.
(b) Calculate Lesedi’s acceleration during the whole motion using the v vs. t graph.
(c) Calculate Lesedi’s displacement during the whole motion using the v vs. t graph.
E - 2. Thandi takes 200 s to walk 100 m to the bus stop every morning. Draw a graph of Thandi’s position as a function of time (assuming that Thandi’s home is the reference point). Use the gradient of the x vs. t graph to draw the graph of velocity vs. time. Use the gradient of the v vs. t graph to draw the graph of acceleration vs. time.
E -3. In the evening Thandi takes 200 s to walk 100 m from the bus stop to her home. Draw a graph of Thandi’s position as a function of time (assuming that Thandi’s home is the origin). Use the gradient of the x vs. t graph to draw the graph of velocity vs. time. Use the gradient of the v vs. t graph to draw the graph of acceleration vs. time.
E - 4. Discuss the differences between the two sets of graphs in questions 2 and 3.

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