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Physics Lesson 3.9.3 - Mathematics of Position vs Time Graph

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Welcome to our Physics lesson on Mathematics of Position vs Time Graph, this is the third 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.

Mathematics of Position vs Time Graph

In the previous tutorial "Equations of Motion", one of the motion equations in the uniformly accelerated (decelerated) motion was

∆x = v0 × t + a × t2/2

Since ∆x = x - x0, we obtain

x - x0 = v0 × t + a × t2/2

Or,

x = x0 + v0 × t + a × t2/2

We can also write the above equation in such a way that the powers of the independent variable decrease from left to right, i.e.

x = a × t2/2 + v0 × t + x0

Time (t) represents the independent variable in this equation while the final position x is the dependent variable. This means this is a quadratic function (as stated in the previous tutorail) because the highest order of the independent variable is two (time is at power two in the second term of the equation).

The mathematical formula of a quadratic function is

y = A × x2 + B × x + C

In the actual application of the above equation in Kinematics, we have

A = a/2, B = v0 and C = x0

From Mathematics, it is known that the graph of a quadratic function is a parabola. When the coefficient A is positive, the arms of the parabola are upwards, and when A is negative, the parabola is arm-down. This means when acceleration is positive, the arms of the parabola are upwards, i.e. the position increases more and more for equal time intervals as seen in the example of the previous paragraph. Look at the graphs below:

Physics Tutorials: This image shows

If we have a negative (arms-down) parabola, the first half of the graph means the object still moves in the positive direction but it is slowing down, until it stops. Then, it turns back, i.e. it starts moving towards negative (although it may still be in the positive part of the position). This motion now is accelerated although the sign of acceleration is negative (this occurs due to the fact that the object is moving towards negative as discussed in the Remark paragraph of the previous tutorial).

Physics Tutorials: This image shows

This is the reason why we take the initial sign of acceleration to describe the entire motion although we may know the object changes direction as in the case of an object thrown upwards at velocity v0.

The table below clarifies the sign of acceleration in different situations:

Sign of acceleration in different situations
How the object is moving?Speeding up towards positiveSlowing down towards positiveSpeeding up towards negativeSlowing down towards negative
What is the sign of acceleration?PositiveNegativeNegativePositive

Let's consider a numerical example.

Example 2

The position vs time graph below belongs to an object thrown upwards. Calculate the initial velocity v0 of this object and the total time of flight t.

Physics Tutorials: This image shows

Solution 2

We have the following information from the graph:

  • hmax = 40m as it is a turning point in the graph. Also, it is known that at hmax the velocity v is zero.
  • tup is the horizontal (time) coordinate of the graph which represents the time needed to reach the turning point.

Thus, using the above info, we obtain

v2 - v20 = 2 × g × hmax

We take g = -10 m/s2 as the object initially is moving up. Therefore, substituting the values, we obtain

02 - v20 = 2 × (-10) × 40
-v20 = -800
v0 = √800
= 28.3 m/s

To find the total time of flight, we use the above result to calculate the time needed for the object to reach the highest point, i.e. first we have to find tup. Then, we have to multiply the result by 2 (by symmetry) to get the total time of flight.

Thus, applying the first equation of vertical motion we have

v = v0 + g × tup
0 = 28.3 + (-10) × tup
10 × tup = 28.3
tup = 2.83 s

Hence:

ttotal = 2 × tup = 2 × 2.83 s = 5.66 s.

We can also find the same result by applying directly the quadratic equation of vertical motion. Thus, the object is in two instants at the height h = 0: one when it starts moving up (at t1 = 0) and the other when it falls again on the ground after flight (at t2). Hence, we have

h = v0 × t + g × t2/2

Substituting the known values, we obtain

0 = 28.3 × t + (-10) × t2/2
0 = 28.3 × t - 5 × t2
0 = t × (28.3 - 5t)

We have

t1 = 0

and

28.3 - 5t2 = 0
5t2 = 28.3
t2 = 28.3/5
= 5.66 s

As you see, both methods give the same results.

You have reach the end of Physics lesson 3.9.3 Mathematics of Position vs Time Graph. 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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