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Physics Lesson 3.9.4 - Distance vs Time Graph in Uniform Motion

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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.

Distance vs Time Graph in Uniform Motion

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:

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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

Physics Tutorials: This image shows

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.

Example 3

An object starting from rest is moving at one direction. Its Position vs Time graph is shown below:

Physics Tutorials: This image shows

Calculate

  1. The average velocity of the entire motion
  2. The average speed of the entire motion

Solution 3

The average velocity < v > is calculated by dividing the total displacement ∆xtot and the total time ttot. From the graph we can see that the object starts moving from x0 = 0 and ends its motion at x = 120m. Again from the graph we can see that the total time of motion is ttot = 25s. Therefore, we obtain for the average velocity:

< v > = ∆xtot/ttot = 120m/25s = 4.8 m/s

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 x0 = 0, goes at x1 = 300m and then turns back by 300m - 120m = 180m because it ends its motion at x2 = 120m. Therefore, the total distance is

stot = 300m + 180m = 480m

Hence, the average speed < v > of the entire motion is

< v > = stot/ttot = 480m/25s = 19.2 m/s

Example 4

The position vs time graph for a moving object is shown below.

Physics Tutorials: This image shows
  1. If the object is initially moving at a certain velocity v0, what is the acceleration of the object?
  2. What is the position x at the end of motion? The magnitude of acceleration is the same during the entire process.
  3. What is the total distance travelled by the object?

Solution 4

From the graph (and problem's clues) we have the following values:

x0 = 12m
x1 = 54m
Δt1 = 6s
Δt2 = 3s
v1 = 0

From the graph, we can calculate the displacement ∆x1 during the first 6 s. We have

∆x1 = x1 - x0
= 54 m - 12 m
= 42 m

Now, we can calculate the initial velocity v0 as it will help us find the acceleration. Thus,

∆x1 = (v0 + v1) × ∆t1/2
42 = (v0 + 0) × 6/2
84 = 6 × v0
v0 = 14 m/s

Now, let's find the acceleration. We have

∆x1 = v0 × ∆t1 + a × t12/2

Substituting the known values, we obtain

42 = 14 × 6 + a × 62/2
42 = 84 + 18 × a
42 - 84 = 18 × a
-42 = 18 × a
a = -42/18 = -7/3m/s2

We can use again the last equation to determine the final position x of the object. The only difference is that this time we have Δt = Δt1 + Δt2 = ttot = t = 9s.

We have

x = x0 + v0 × t + a × t2/2

Substituting the known values, we obtain

x = 12 + 14 × 9 + (-7/3) × 92/2
= 12 + 126 - 7 × 81/6
= 138 - 94.5= 43.5m

Therefore, the object will be at x = 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,

∆x2 = x - x1
= 43.5m - 54m
= -10.5m

This means s2 = 10.5m. Also, we have s1 = ∆x1 = 42m. Therefore, the total distance s is

s = s1 + s2
= 42m + 10.5m
= 52.5m

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.

More Position v's Time and Distance v's Time Graph Lessons and Learning Resources

Kinematics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
3.9Position v's Time and Distance v's Time Graph
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
3.9.1Position verse Time Graph in Uniform Motion
3.9.2Position vs Time Graph in Uniformly Accelerated (Decelerated) Motion
3.9.3Mathematics of Position vs Time Graph
3.9.4Distance vs Time Graph in Uniform Motion

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  6. Continuing learning kinematics - read our next physics tutorial: Velocity v's Time and Speed v's Time Graph

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