# Physics Tutorial: Position v's Time and Distance v's Time Graph

In this Physics tutorial, you will learn:

• How to interpret a position vs time graph in uniform motion?
• The same for uniformly accelerated (decelerated) motion
• To understand the difference between position vs time and distance vs time graph
• How to convert the abovementioned graphs from one type into the other
• Understand the concept of "gradient" in a position vs time graph and what it represents
• Confirm the equations of motion through the position vs time graph
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3.9Position v's Time and Distance v's Time Graph

## Introduction

In the Physics tutorial: "Speed and Velocity in One Direction" we discussed about the procedure of finding the instantaneous speed and instantaneous velocity and we saw that position vs time graphs or distance vs time graphs were very helpful in this regard.

Now, we will discuss more in detail this method by extending the explanation in other aspects as well. Another advantage of motion graphs in general, is that when dealing with them, the veracity of the equations of motion is confirmed.

## Position vs Time Graph in Uniform Motion

As discussed in the tutorial "Speed and Velocity in One Direction", the Position vs Time graph consists in two perpendicular axes where Time is shown in the horizontal axis and Position in the vertical one as shown below: As you see, the position axis lies both in positive and negative part of coordinate as the position can be also negative but the time axis lies only due positive as no negative time exists.

It is very important to emphasize that the graph does not show the trajectory of the object; it only provides information on how the object moves. That mean if a Position vs Time graph for a uniform motion is like this, It does not mean the object is ascending an uphill; rather, it only means the object is moving linearly at constant velocity, as at equal time intervals, the position of the object changes uniformly. This conclusion is drawn by considering the concept of "gradient" in Math. It is known that Gradient of a slope is obtained by dividing the change in the vertical coordinate (here the change in position or displacement) to the change in the horizontal one (here the "change in time" or the "time interval involved") as shown below: From the above reasoning, it is clear that:

• "The gradient of the Position vs Time graph at any point of the graph gives the Instantaneous Velocity".
• If the motion is uniform as in the example shown in the graphs above, the instantaneous velocity is the same everywhere; it is the equal to the average velocity. Therefore, we can write:
< v > = v = ∆x/∆t

### Example 1

An object starts moving from x0 = -12 m and it goes at x1 = 9m in 7 seconds. After reaching this position it stops for 3 seconds and then moves for other 5 s until it reaches the position x2 = 4m.

• Plot the position vs time graph for each interval
• Calculate the velocity in each interval

### Solution 1

In the first 7 s the object moves from x = -12 m to x = 9 m. Then it stays at x = 9 m for other 3 s (from 7s to 10 s). Finally, it moves from x = 9 m to x = 4m during the last 5s (from 10s to 15s). The position vs time graph is shown below: The velocity for each interval is

v1 = x1-x0/t1
= 9m-(-12m)/7s
= 21m/7s
= 3 m/s

In the second interval there is no motion (x1 = x2 = 9m), so the velocity is zero because

v1 = x2 - x1/t2
= 9m - 9m/3s
= 0

In the third interval, we have

v3 = x3-x2/t3
= 9m - 4m/5s
= 5m/5s
= 1 m/s

## Position vs Time Graph in Uniformly Accelerated (Decelerated) Motion

In non-uniform motion, the position vs time graph is not linear as the position does not change at the same rate everywhere as shown in the graph below. From the above graph, it can be seen that the displacement Δx2 in the second interval is greater than the displacement Δx1 in the first one. As a result, the velocity v2 in the second interval is greater than the velocity v1 in the first one because the intervals Δt are equal. Mathematically, we have

∆x2 > ∆x1∆x2/∆t > ∆x1/∆t ⟹ v2 > v1

Therefore, from the above discussion we can draw the following conclusion:

"Steeper the slope (greater the gradient) of a position vs time graph, greater the moving velocity of the object."

In this kind of motion, the slope changes at every instant, so we can only calculate the instantaneous velocity through the abovementioned method, not the average velocity. Look at the figure: It is obvious that slope 2 is steeper than the slope 1. This means the velocity v2 at the point where the slope 2 touches the graph is greater than the velocity v1 at the point where the slope 1 touches the graph.

## 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: 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). 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. ### 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.

## 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: 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.

### Example 3

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

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