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Physics Lesson 3.10.3 - Velocity verse Time Graph in Uniformly Accelerated (Decelerated) Motion
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Welcome to our Physics lesson on Velocity verse Time Graph in Uniformly Accelerated (Decelerated) Motion, this is the third lesson of our suite of physics lessons covering the topic of Velocity v's Time and Speed v's Time Graph, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
Velocity verse Time Graph in Uniformly Accelerated (Decelerated) Motion
In the previous tutorials, we stated that in uniformly accelerated (decelerated) motion, i.e. in the motion with constant acceleration, the velocity increases or decreases at the same rate. This means the Velocity vs Time graph of this kind of motion will be a sloped straight line as shown below
In the first part of the graph, the acceleration is negative as the velocity decreases, while in the second part, the acceleration is positive as the velocity increases.
Also, here again we can find the displacement by calculating the area under the graph as stated in the previous paragraph. Thus, the area bordered by the graph and the horizontal axis gives the numerical value of the displacement. Let's explain this point by illustrating it with numbers.
Example 1
The velocity vs time graph below represents the motion of an object in a linear direction.
Calculate:
- The acceleration during the first 10 seconds
- The acceleration during the next 15 seconds
- The total displacement of the object
Solution 1
From the graph, we can extract the following values for the first interval. We have
v⃗1 = 4 m/s
Δt1 = 10 s
Thus, we obtain
= 4 m/s - 15 m/s/10 s
= -11 m/s/10 s
= -1.1 m/s2
The same procedure is used to calculate the acceleration during the second interval Δt2. We have
v⃗ = 10 m/s
Δt2 = 25 s - 10 s = 15 s
Therefore,
= 10 m/s - 4 m/s/15 s
= 6 m/s/15 s
= 0.4 m/s2
We have to calculate the area under the graph to calculate the displacement. Both parts of the graph represent trapeziums, so their area is calculated by the equation
In the first interval, we have:
Shorter base = v⃗1 = 4m/s
Height = ∆t1 = 10 s
Area 1 = Displacement 1 = ∆x⃗1
Hence,
= (15 m/s + 4 m/s) × 10s/2
= 95 m
The same procedure is used in the second interval as well. We have:
Shorter base = v⃗1 = 4 m/s
Height = ∆t2 = 15 s
Area 2 = Displacement 2 = ∆x⃗2
Thus,
= (10 m/s + 4 m/s) × 15s/2
= 105 m
Therefore, the total displacement is
= 95 m + 105 m
= 200 m
In the above problem, it was confirmed the veracity of the second formula of the uniformly accelerated (decelerated) motion
You have reach the end of Physics lesson 3.10.3 Velocity verse Time Graph in Uniformly Accelerated (Decelerated) Motion. There are 4 lessons in this physics tutorial covering Velocity v's Time and Speed v's Time Graph, you can access all the lessons from this tutorial below.
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