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In this Physics tutorial, you will learn:

- The properties of Velocity vs Time graph in the uniform motion
- How to plot the corresponding Speed vs Time graph when the Velocity vs Time graph is given?
- The meaning of the area under the velocity or speed vs time graph
- The same for the uniformly acceleration (decelerated) motion
- How to plot the Position vs Time graph when the corresponding Velocity vs Time graph is given?

Tutorial ID | Title | Tutorial | Video Tutorial | Revision Notes | Revision Questions | |
---|---|---|---|---|---|---|

3.10 | Velocity v's Time and Speed v's Time Graph |

As discussed in the previous Physics tutorial "Position vs Time & Distance vs Time Graph", we obtain useful information from motion graphs. They represent another "language" by which Physics communicates with its readers besides the language of words. Therefore, motion graphs are very important when trying to find the relationship between physical quantities involved in Kinematics but not only.

In this tutorial, we will deal with Velocity vs Time and Speed vs Time graphs. The approach will be very gradual, so the reader will have the chance to understand in depth these kind of motion graphs.

Like in the Position vs Time graph, in the Velocity vs Time graph the horizontal axis contains the Time, t, while the velocity is shown at the vertical axis.

In uniform motion, the velocity is constant. This means the Velocity vs Time graph will be a horizontal line, which lies v*⃗* units above or below zero depending on the sign of velocity. This line extends horizontally from zero up to the moving time t.

Let's explain this point by illustrating it with numbers. Suppose we are required to find the displacement and distance of the object whose motion is shown in the graph below.

From the graph we can see that the velocity during the first 10 s is constant and positive (v*⃗*_{1} = 12 m/s). This means during this interval the object is moving at constant velocity in the positive direction of the position axis.

During the next 15 seconds, the object has a negative constant velocity (v*⃗*_{2} = -4 m/s). This means it is moving at constant velocity towards the negative direction of position axis (although it may still be moving on the positive part of the position axis).

Therefore, applying the equation of uniform motion for both parts of the motion, we obtain

∆x*⃗*_{1} = v*⃗*_{1} × t_{1}=12 *m**/**s* × 10 s = 120 m

and

∆x*⃗*_{2} = v*⃗*_{2} × t_{2} = -4 *m**/**s* × 15 s = -60 m

Hence, the total displacement ∆x*⃗*_{tot} is

∆x*⃗*_{tot} = ∆x*⃗*_{1} + ∆x*⃗*_{2}

= 120 m + (-60)m

= 60 m

= 120 m + (-60)m

= 60 m

As we have explained in the tutorial "Speed and Velocity in One Direction", the speed cannot be negative (and time as well). Therefore, the Speed vs Time graph will contain only the upper-right part of a two dimensional graph. Let's explain this point by considering the numerical values of the previous example. Any negative part of the Velocity vs Time graph is flipped vertically and it will be considered as positive in the corresponding Speed vs Time graph as shown below.

The total Distance s_{tot} will therefore be:

s_{tot} = s_{1} + s_{2}

= v_{1} × t_{1} + v_{2} × t_{2}

= 12*m**/**s* × 10 s + 4 *m**/**s* × 15 s

= 120 m + 60 m

= 180 m

= v

= 12

= 120 m + 60 m

= 180 m

The above results comply with the numerical values of the area enclosed by the graph and the horizontal axis. If the graph in the velocity vs time graph is in the negative part, the area is taken as negative. Therefore, we obtain a very important property of the Velocity vs Time and Speed vs Time graphs:

**The area under (or above when negative) of the Velocity vs Time graph gives the Displacement**, and**The area under the Speed vs Time graph gives the Distance.**

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.

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

From the graph, we can extract the following values for the first interval. We have

v*⃗*0 = 15 m/s

v*⃗*1 = 4 m/s

Δt1 = 10 s

v

Δt1 = 10 s

Thus, we obtain

a*⃗*_{1} = *v**⃗*_{1}-v*⃗*_{0}*/**∆t*_{1}

=*4 **m**/**s* - 15 *m**/**s**/**10 s*

=*-11 **m**/**s**/**10 s*

= -1.1*m**/**s*^{2}

=

=

= -1.1

The same procedure is used to calculate the acceleration during the second interval Δt_{2}. We have

v*⃗*1 = 4 m/s

v*⃗* = 10 m/s

Δt2 = 25 s - 10 s = 15 s

v

Δt2 = 25 s - 10 s = 15 s

Therefore,

a*⃗*_{2} = *v**⃗* - v*⃗*_{1}*/**∆t*_{2}

=*10 **m**/**s* - 4 *m**/**s**/**15 s*

=*6 **m**/**s**/**15 s*

= 0.4*m**/**s*^{2}

=

=

= 0.4

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

Area of trapezium = *(Longer base + Shorter base) × Height**/**2*

In the first interval, we have:

Longer base = v*⃗*_{0} = 15 *m**/**s*

Shorter base = v*⃗*_{1} = 4*m**/**s*

Height = ∆t_{1} = 10 s

Area 1 = Displacement 1 = ∆x*⃗*_{1}

Shorter base = v

Height = ∆t

Area 1 = Displacement 1 = ∆x

Hence,

∆x*⃗*_{1} = *(v**⃗*_{0} + v*⃗*_{1} ) × ∆t_{1}*/**2*

=*(15 **m**/**s* + 4 *m**/**s*) × 10s*/**2*

= 95 m

=

= 95 m

The same procedure is used in the second interval as well. We have:

Longer base = v*⃗* = 10 m/s

Shorter base = v*⃗*_{1} = 4 m/s

Height = ∆t_{2} = 15 s

Area 2 = Displacement 2 = ∆x*⃗*_{2}

Shorter base = v

Height = ∆t

Area 2 = Displacement 2 = ∆x

Thus,

∆x*⃗*_{2} = *(v**⃗* + v*⃗*_{1} ) × ∆t_{2}*/**2*

=*(10 **m**/**s* + 4 *m**/**s*) × 15s*/**2*

= 105 m

=

= 105 m

Therefore, the total displacement is

∆x*⃗*_{tot} = ∆x*⃗*_{1} + ∆x*⃗*_{2}

= 95 m + 105 m

= 200 m

= 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

∆x*⃗* = *(v**⃗*_{0} + v*⃗* ) × ∆t*/**2*

Like in the distance vs time graph, the speed vs time graph is similar to the velocity vs time graph when the acceleration (slope) is positive. On the other hand, in the part of the graph in which the slope is negative (where the acceleration is negative), the graph is obtained by flipping vertically the corresponding section of the velocity vs time graph, as shown below.

The second graph (the speed vs time graph) does not provide any information on the direction of motion; it only shows that the object first decelerates until it stops, then it accelerates. This may either occur when the object changes direction or when it only slows down, stops and accelerates again at the same direction. Therefore, the velocity vs time graph is more comprehensive than the corresponding speed vs time graph.

The velocity vs time graph for a moving object is shown in the figure below.

- Calculate the acceleration in the first 4 seconds
- Calculate the acceleration in the interval (4s - 8s)
- Calculate the acceleration in the interval (8s - 14s)
- Calculate the displacement in the first 4 seconds
- Calculate the displacement in the interval (4s - 8s)
- Calculate the displacement in the interval (8s - 14s)
- Calculate the total displacement of the object
- Plot the corresponding speed vs time graph for the entire motion

From the graph, we can extract the following values during the first 4 s:

v*⃗*_{0} = - 6 *m**/**s*

v*⃗*_{1} = 0

t1 = 4s

v

t1 = 4s

Thus, we have

a*⃗*_{1} = *v**⃗*_{1} - v*⃗*_{0}*/**t*_{1}= *0 - (-6) **m**/**s**/**4 s*= *6m/s**/**4 s*= 1.5 m/s

As you see, the acceleration is positive although the object is slowing down. This is because it is moving in the negative part of coordinates and slowing down in the negative part of coordinates has the same effect as speeding up in the positive one.

In the following 4 s (from 4s to 8s) we have:

v*⃗*1 = 0

v*⃗*_{2} = 7 *m**/**s*

t2 = 8s - 4s = 4s

v

t2 = 8s - 4s = 4s

Therefore,

a*⃗*_{2} = *v**⃗*_{2} - v*⃗*_{1}*/**t*_{2} = *7 **m**/**s* - 0 *m**/**s**/**4 s* = *7 m/s**/**4 s * = 1.75 m/s^{2}

In the third interval (from 8s to 14s) we have:

v*⃗*_{2} = 7 m/s

v*⃗*_{3} = 3 m/s

t3 = 14 s - 8 s = 6 s

v

t3 = 14 s - 8 s = 6 s

Hence, we have

a*⃗*_{3} = *v**⃗*_{3} - v*⃗*_{2}*/**t*_{3} = *3 **m**/**s* - 7 *m**/**s**/**6 s* = *-4 m/s**/**6 s* = -0.67 m/s

The displacement in the first 4s is obtained by calculating the magnitude of the corresponding area of the figure, i.e.

∆x*⃗*_{1} = *(v**⃗*_{1}+v*⃗*_{0}) × t_{1}*/**2* = *(-6 +0) × 4**/**2* = -12m

The displacement in the next 4s (from 4s to 8s) is obtained by calculating the magnitude of the corresponding area of the figure, i.e.

∆x*⃗*_{2} = *(v**⃗*_{2} + v*⃗*_{1}) × t_{2}*/**2* = *(7 + 0) × 4**/**2* = 14m

The displacement in the last 6s (from 8s to 14s) is obtained by calculating the corresponding area of figure, i.e.

∆x*⃗*_{3} = *(v**⃗*_{3}+v*⃗*_{2}) × t_{3}*/**2* = *(3 + 7) × 6**/**2* = 30m

The total displacement therefore, is obtained by adding all the previous displacements

∆x*⃗*_{tot} = ∆x*⃗*_{1} + ∆x*⃗*_{2} + ∆x*⃗*_{3}

= -12 m + 14m +30m

= 32m

= -12 m + 14m +30m

= 32m

The speed vs time graph differs from the velocity vs time graph only in the negative part (the graph is flipped vertically). Thus,

How to plot the Position vs Time graph when the corresponding Velocity vs Time graph is given?

First, let's take a simple velocity vs time graph to explain this idea and then we will deal with more complicated graphs.

An object starts moving from x*⃗*_{0} = 12 m. Its velocity vs time graph is shown below.

Plot the corresponding Position vs Time graph for this motion.

The graph shows a uniformly accelerated motion where v*⃗*_{0} = 4 m/s, v*⃗* = 16 m/s, t = 6s and x*⃗*_{0} = 12m. From these clues, we can calculate the acceleration a*⃗* for this motion. Thus,

a*⃗* = *v**⃗* - v*⃗*_{0}*/**t* = *16 **m**/**s* - 4 *m**/**s**/**6 s* = 2 *m**/**s*^{2}

Thus giving that the equation of motion with constant acceleration is a quadratic equation of the type

y(x) = A × x^{2} + B × x + C

where

A = *a**⃗**/**2* = *2**/**2* = 1

B = v*⃗*_{0} = 4

C = x*⃗*_{0} = 12

x = t (the independent variable)

B = v

C = x

x = t (the independent variable)

we obtain after the substitutions

x(t) = 1 × t^{2} + 4 × t + 12

Or

x(t) = t^{2} + 4t + 12

The graph of this quadratic function is a parabola. Therefore, we need to find several values in order to plot an accurate graph. These values are obtained by substituting t by 0, 1, 2, 3, 4, 5 and 6 respectively. The following table shows the results obtained.

Time t (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|---|

Position x (m) | 12 | 17 | 24 | 33 | 44 | 57 | 72 |

Now, let's insert these points in the position vs time graph and connect them smoothly.

As you see, the graph is a positive (arms-up) parabola since the acceleration is positive (the object speeds up).

Now, let's consider a more complicated example in which multiple kinds of motion are involved.

Plot the position vs time graph for the motion described in the velocity vs time graph shown below if initially the object was at x*⃗*_{0} = 30 m.

From the graph we can see the motion is composed by three parts: the first part (from 0 to 12 s) represents an accelerated motion; in the second part (from 12 s to 22 s) the motion is uniform; and in the third part (from 22 s to 30 s) the motion is decelerated (by the end of which the object eventually stops because the final velocity becomes zero).

a. In the first interval the approach is similar to that of the previous example. We have v*⃗*_{0} = 5 m/s, v*⃗*_{1} = 8 m/s, t_{1} = 12s. Therefore, we have for the acceleration a*⃗*_{1}:

a*⃗*_{1} = *v**⃗*_{1} - v*⃗*_{0}*/**∆t*_{1} = *8 **m**/**s* - 5 *m**/**s**/**12 s* = 0.25 *m**/**s*^{2}

Therefore, the equation of motion for the first interval is

x*⃗*_{1} = x*⃗*_{0} + v*⃗*_{0} × ∆t_{1} + *a**⃗*_{1} × ∆t^{2}_{1}*/**2*

Substituting the values, we obtain

x*⃗*_{1} = 30 + 5 × 12 + *0.25 × 12*^{2}*/**2*

= 108 m

= 108 m

We can find some points in the graph to make the graph more accurate. We have

t (s) | 3 | 6 | 9 | 12 |
---|---|---|---|---|

x⃗ = 30 + 5 × t + 0.25 × t^{2}/2 | 46.125 m | 64.5 m | 85.125 m | 108 m |

We will insert later these points in the final graph.

b. In the second interval, the velocity is constant (v*⃗*_{1} = v*⃗*_{2} = 8 ** m/s**). Therefore, we have

∆x*⃗*_{2} = x*⃗*_{2} - x*⃗*_{1}

= v*⃗*_{1} × ∆t_{2}

= 8*m**/**s* × (22s - 12s)

= 8*m**/**s* × 10s

= 80 m

= v

= 8

= 8

= 80 m

This means by the end of the second interval the object will be at

x*⃗*_{2} = x*⃗*_{1} + ∆x*⃗*_{2}

= 108 m + 80 m

= 188 m

= 108 m + 80 m

= 188 m

c. In the third interval, the motion is decelerated. The approach is the same as in the first part. The only difference is that here the initial position is x*⃗*_{2} = 188 m.

We have

a*⃗*_{3} = *v**⃗*_{3} - v*⃗*_{2}*/**∆t*_{3} = *0 **m**/**s* - 8 *m**/**s**/**30 s - 22 s* = *-8 **m**/**s**/**8 s* = -1 m/s^{2}

The general equation of the third interval is

x*⃗*_{3} = x*⃗*_{2} + v*⃗*_{2} × ∆t_{3} + *a**⃗*_{3} × ∆t^{2}_{3}*/**2*

Substituting the known values, we obtain the specific equation of motion in the third interval,

x*⃗*_{3} = 188 + 8 × ∆t_{3} + *(-1) × ∆t*^{2}_{3}*/**2*

or

x*⃗*_{3} = 188 + 8 × ∆t_{3} - 0.5 × ∆t^{2}_{3}

Hence, at the end of the motion the object will be at

x*⃗*_{3} = 188 + 8 × 8 - 0.5 × 8^{2}

= 188 + 64 - 32

= 220 m

= 188 + 64 - 32

= 220 m

Again, we can use the table method to find some intermediate points in the graph to make its plotting more accurate. We have

t (s) | 2 (24 - 22) | 4 (26 - 22) | 6 (28 - 22) | 8 (30 - 22) |
---|---|---|---|---|

x⃗ = 188 + 8 × t - 0.5 × t^{2} | 202 m | 212 m | 218 m | 220 m |

Now, let's insert all the above values obtained during the solution of this exercise in the same Position vs Time graph

The Distance vs Time graph would be the same as the Velocity vs Time in this example as the entire movement has occurred only in the positive direction.

Enjoy the "Velocity v's Time and Speed v's Time Graph" physics tutorial? People who liked the "Velocity v's Time and Speed v's Time Graph" tutorial found the following resources useful:

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- Continuing learning kinematics - read our next physics tutorial: Acceleration v's Time Graph

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