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Welcome to our Physics lesson on Distance vs Time Graph in Uniform Motion, this is the fourth lesson of our suite of physics lessons covering the topic of Position v's Time and Distance v's Time Graph, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
The Distance vs Time graph is identical to the Position vs Time graph when the object is moving in the positive direction but it is flipped vertically when the object is moving towards negative. This is because distance cannot be negative; it represents the path length followed by an object during its motion. The other difference is that unlike in the Position vs Time graph, the vertical axis of the Distance vs Time graph cannot extend towards the negative direction (Distance cannot be negative). Look at the figures below:
Figure (a) shown above: A Displacement vs Time graph for a uniform motion (in two parts) and the corresponding Distance vs Time graph for the same motion
Figure (b) shown above: A Displacement vs Time graph for a motion with uniform acceleration (in two parts) and the corresponding Distance vs Time graph for the same motion. As you may see, the object stops for a while at the end of the first interval and starts moving again at the beginning of the second interval.
An object starting from rest is moving at one direction. Its Position vs Time graph is shown below:
Calculate
The average velocity < v⃗ > is calculated by dividing the total displacement ∆x⃗tot and the total time ttot. From the graph we can see that the object starts moving from x⃗0 = 0 and ends its motion at x⃗ = 120m. Again from the graph we can see that the total time of motion is ttot = 25s. Therefore, we obtain for the average velocity:
The average speed < v > is calculated by dividing the total distance stot and the total time ttot. From the graph we can see that the object starts moving from x⃗0 = 0, goes at x⃗1 = 300m and then turns back by 300m - 120m = 180m because it ends its motion at x⃗2 = 120m. Therefore, the total distance is
Hence, the average speed < v > of the entire motion is
The position vs time graph for a moving object is shown below.
From the graph (and problem's clues) we have the following values:
From the graph, we can calculate the displacement ∆x⃗1 during the first 6 s. We have
Now, we can calculate the initial velocity v⃗0 as it will help us find the acceleration. Thus,
Now, let's find the acceleration. We have
Substituting the known values, we obtain
We can use again the last equation to determine the final position x⃗ of the object. The only difference is that this time we have Δt = Δt1 + Δt2 = ttot = t = 9s.
We have
Substituting the known values, we obtain
Therefore, the object will be at x⃗ = 43.5m at the end of its motion.
In order to find the total distance, we must calculate the second displacement and then adding it to the first one. We must take only positive values here. Thus,
This means s2 = 10.5m. Also, we have s1 = ∆x⃗1 = 42m. Therefore, the total distance s is
You have reach the end of Physics lesson 3.9.4 Distance vs Time Graph in Uniform Motion. There are 4 lessons in this physics tutorial covering Position v's Time and Distance v's Time Graph, you can access all the lessons from this tutorial below.
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