# MUV time diagram

When a body changes its velocity in a constant way, we say that it is in uniformly varied motion (MUV).

**Uniformly**

**varied**

**motion**(MUV) hourly diagrams are graphs that relate quantities, such as

**position**,

**velocity**, and

**acceleration,**with the passage of

**time.**Using the MUV diagrams , we can calculate the

**speed**of the mobile, its

**displacement**and even its speed

**variations**

**.**

**Velocity by time (vxt) graph in MUV**

The graph of velocity versus time in uniformly varied motion is always an **ascending** or **descending ****line** , since velocity in this type of motion is subject to an acceleration or **deceleration** of constant magnitude.

When the velocity of a mobile increases as a function of time, it is represented as an ascending line. This motion is called accelerated motion **.**

The formula used to plot v(t) (velocity versus time) is called the hourly function of velocity. This function is shown below:

**Caption** :

**v **_{f} – final velocity (m/s)

**v **_{0} – initial velocity (m/s)

**a** – average acceleration (m/s²)

**t** – time interval (s)

The above function is a 1st degree (straight) function. Therefore, for **positive ****values** of **acceleration,** your graph will be an **ascending ****line** , as shown below:

If the velocity of the object decreases as a function of time, we say that its **motion** is retarded **. **In this case, the velocity versus time diagram will also be a straight line, but that line will descend because of the **negative ****acceleration .**

In addition, this type of graph is especially useful for calculating the displacement suffered by the mobile. To do this, just calculate the area of this graph.

In the graph above, although the velocity module is decreasing, it is still located above the abscissa axis (time axis). This indicates that the mobile is still moving away from the origin. It is, therefore, a **delayed progressive ****movement .**

**Position graph by time (S xt) in MUV**

The position versus time diagram for the MUV is described by the graph of a **second ****degree** function called the **hourly ****function ****of ****the ****MUV ****position . **This function can be written in two ways **:**

**Legend:
S **

_{f}

**–**final position (m)

**S**

_{0 –}initial position (m)

**ΔS = S**

_{f}

**– S**

_{0}

**–**displacement (m)

**v**

_{f}– final velocity (m/s)

**v**

_{0}– initial velocity (m/s)

**a**– average acceleration (m/s²)

**t**– time interval (s)

The position hour function for the **MUV** indicates that the **position (S)** versus **time (t)** graph will have the shape of a parabola. When the speed of the mobile increases as a function of time, we will have a **parabola** with the concavity **facing ****upwards ****. **Watch:

In cases where the motion **slows down,** we will have a graph of a **parabola** with the **concavity ****facing ****downwards ****. **Watch:

The most important points of position versus time diagrams are their roots (t’ and t” or t _{0} and t _{f} ). These points inform us when the mobile passes through the origin of the reference frame. Furthermore, for the initial time instant (t _{0} ), it is possible to determine the **initial ****position** (S _{0} ) of the mobile: just look at how high the parabola crosses the vertical axis during the initial time instant.

**Graph of acceleration as a function of time (axt) in the MUV**

Graphs of acceleration versus time are especially useful in determining whether a body is being **accelerated** or **decelerated. **By means of these graphs, it is possible to calculate the variation of the velocity suffered by the mobile (Δv = v _{f} – v _{0} ).

For uniformly accelerated motion, we will have a line parallel to the horizontal axis of constant height. Watch:

Since the acceleration line is above the horizontal axis, we say that the motion is accelerated. To determine the change in the speed of the mobile, just calculate the **area ****of this ****graph.**

When the mobile is undergoing some **deceleration** , the line **a(t)** (acceleration as a function of time) will appear **below** the horizontal axis, since the values of this line will be negative. Watch:

**Summary about the MUV time chart**

→ In uniformly **accelerated motion , the ****position** versus **time** graph s(t) will be a parabola with the **concavity ****facing ****upwards ****,** and its **acceleration** will be **positive.**

→ In uniformly **decelerated** or retarded motion, the **position** versus **time** graph s(t) will be a parabola with the **concavity ****facing ****downwards ****,** and its **acceleration** will be **negative.**

→ In the uniformly **accelerated** motion , the graph of velocity versus time **v(t)** will be an **ascending** (rising) line.

→ In uniformly **retarded** motion , the graph of velocity versus time **v(t)** will be a **descending** (going down) line.

→ In uniformly **accelerated** motion , the graph of acceleration as a function of time **a(t)** will be a line **parallel to the ****horizontal** axis and arranged **above** it.

→ In uniformly **retarded** motion , the graph of acceleration as a function of time **a(t)** will be a line **parallel to the ****horizontal** axis and lying **below** it.

**Solved exercises on MUV graphics**

1. The graph of **velocity** versus time of a mobile moving with constant acceleration is shown below:

Analyze the graph above and answer:

a) What is the initial velocity of the car?

b) What is the final velocity of the car?

c) What is the magnitude of the car’s acceleration?

d) Classify the movement as progressive or regressive, retarded or accelerated.

e) Calculate, through the graph, the magnitude of the displacement suffered by the mobile.

f) Determine at which instant the speed of the mobile will have a magnitude of 60 m/s.

**Resolution:**

**a)** Analyzing the graph above, it is possible to notice that, at the initial instant of time (t _{0} = 0 s), where the two axes of the graph cross, the speed of the mobile is **20 m/s** .

**b)** The final time instant (t _{f} ) shown in the graph is the instant t = 5 s. At that instant of time, the speed of the mobile is **50 m/s** .

**c)** We can calculate the acceleration of the mobile through the equation of the average acceleration (equivalent to the hourly velocity function):

Analyzing the graph, it is possible to find the values of **v **_{f} , **v **_{i}** , ****t **_{f} and **t **_{i. }That way:

The calculation indicates that the speed of the mobile increases, in module, **6 meters per second every second** .

**d)** The velocity line is above the horizontal axis, therefore, the movement is progressive, that is, the mobile moves away from the referential. Also, the line is sloping upward, indicating that its magnitude increases. It is, therefore, an accelerated movement. Therefore, we say that the motion is **progressive** and **accelerated** .

**e)** To calculate the distance traveled by the mobile through the graph, we must calculate the area. The graph has the geometric shape of a trapezoid, whose area is given by the formula:

**Legend:
A** – area of the trapezoid

**B**– edge of the larger base

**b**– edge of the smaller base

**h**– height of the trapeze

We can assimilate the velocity **v **_{f} to the major base **B,** the initial velocity **vi** to the minor base _{b} , the final instant of time **t ****f** to the height _{h}** of the trapezoid** . With that, we will have:

**f)** To determine the speed of the mobile at some time later than those shown in the graph, we use the **hourly ****speed ****function ****:**

As calculated earlier, the acceleration of this mobile is equal to 6 m/s². In order for it to reach a final speed of **60 m/s,** we will have the following resolution:

Therefore, we can conclude that the mobile will have a speed of **60 m/s** at the instant of time **t = 10 s** .