# Physics Tutorial: Equations of Motion

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In this Physics tutorial, you will learn:

• Which equations are used in the uniform and non-uniformly accelerated (decelerated) motion?
• How these equations transform when objects move vertically?
• How does free fall differ from vertical motion?
• How to apply the equations of motion in specific situations?
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3.8Equations of Motion

## Introduction

In the last tutorial "The Meaning of Acceleration. Constant and non-Constant Acceleration", we discussed the concept of acceleration and also we obtained the equation of acceleration

### Equation 1

a = ∆v/∆t = v - v0/∆t

where v is the final velocity (the velocity at the end of motion), v0 the initial velocity and Δt the time interval during which the process occurs.

It is worth mentioning that the above formula is valid only for the uniformly accelerated (decelerated) motion, i.e. for the motion in which the acceleration (deceleration) is constant during the entire process.

Also, when dealing with the uniform motion ("Speed and Velocity in One Dimension"), we learned the (sole) equation of this kind of motion

### Equation 2

a = ∆x/∆t = x - x0/∆t

where x is the final position of the object and x0 its initial position (therefore, ∆x is the change in position or displacement).

The above equations deal only with vector quantities (displacement, velocity, acceleration). However, it is clear that the same equations can be used for the corresponding scalar quantities (distance and speed) as well. We will discuss all these physical quantities in the following of this tutorial.

## Equations of the Uniform Motion

In the physics tutoral: "Speed and Velocity in One Dimension" we learned two equations in total. They are:

v = ∆x/∆t = x - x0/∆t

for the vector quantities involved, and

v = s/t

for the corresponding scalar quantities. These are the only equations for the uniform motion (the motion with constant velocity or speed). These equations are very similar and when the object is moving linearly without turning back, numerically they both give the same result.

For example, if an object is moving due left at 6 m/s for 15 s, the magnitude of displacement and distance are both 90 m because both the velocity and speed are 6 m/s. Thus, giving that the time interval is 15 s, we have

v = ∆x/∆t ⇒ ∆x = v × ∆t = 6 m/s × 15 s = 90 m

and

v = s/t ⇒ s = v × t = 6 m/s × 15 s = 90 m

Remark! The total time t and the time interval Δt in most cases represent the same thing because the initial time (ti or t0) usually is taken as zero. This means we start measuring the quantities involved in the process such as position, velocity or speed when we press the stopwatch. Therefore, we have

∆t = tfinal - tinitial
= t - 0
= t

Thus, Δt and t henceforth will represent the total time of motion.

The following table summarizes all said above.

Type of quantities involved Equation Vector Scalar v⃗ = ∆x⃗/∆t v = s/t

## Equations of the Uniformly Accelerated (Decelerated) Motion

From the equation (1), we can derive the first equation of the motion with constant acceleration. Thus, rearranging the equation (1)

a = ∆v/∆t = v - v0/∆t

we obtain

v - v0 = a × ∆t

and finally, we get after sending the initial velocity v0 on the right of the equality sign

### Equation i

v = v0 + a × ∆t

The equation (i) gives the standard form of the first equation of motion with constant acceleration.

Now, let's derive another equation of motion with constant acceleration. When the acceleration is constant, the average velocity can be obtained by calculating the arithmetic mean of the final and initial velocities. Thus, we can write

### Equation 3

< v > = v + v0/2

This averagization helps us calculate much easier the displacement by considering the motion as uniform and using the average velocity written in the above formula as (constant) velocity of a uniform motion. Then, we multiply this average velocity by time to calculate the displacement.

Mathematically, we have:

∆x = < v > × ∆t

Therefore, we obtain after substituting the average velocity < v > with the corresponding expression obtained at

### Equation ii

∆x = v + v0/2 × ∆t

The equation (ii) is the second equation of motion with constant acceleration.

If we express the equation (ii) in another form, i.e. as

(v+v0) = 2 × ∆x/∆t

and multiplying both sides of this equation by (v-v0), we obtain

(v-v0) × (v+v0) = (v-v0) × 2 × ∆x/∆t

Substituting the expression (v - v0) in the right part of the above equation with a × ∆t (which is derived from the equation I), we obtain

(v-v0) × (v+v0) = a × ∆t × 2 × ∆x/∆t

After simplifying Δt from the right side of the above equation, we obtain the third equation of motion with constant acceleration

(v-v0) × (v+v0) = 2 × a × ∆x

or

### Equation iii

v2 - v20 = 2 × a × ∆x

[Here we made use of the well-known algebraic identity (x-y) × (x+y)=x2-y2]

Now, let's derive the fourth (and last) equation of the uniformly accelerated (decelerated) motion. Thus, in the equation (ii), we substitute v by its corresponding expression obtained in the equation (i), i.e. v0 + a × ∆t and in this way, we obtain

∆x = v+v0/2 × ∆t
= (v0 + a × ∆t)+v0/2 × ∆t
= 2v0 + a × ∆t/2 × ∆t
= 2v0 + a × ∆t/2 × ∆t
= 2v0/2 + a × ∆t/2 × ∆t
= v0 × ∆t + a × ∆t2/2

Thus, the fourth and last equation of the uniformly accelerated (decelerated) motion or as it is otherwise known, the motion with constant acceleration, is

### Equation iv

∆x = v0 × ∆t + a × ∆t2/2

The above equations are valid for the scalar quantities involved in the uniformly accelerated (decelerated) motion as well. Thus, using the corresponding quantities and symbols, we write

v = v0 + a × t

### Equation ii

s = ( v + v0 )/2 × t

### Equation iii

v2 - v20 = 2 × a × s

### Equation iv

s = v0 × t + a × t2/2

### Example 1

A car moving initially at 4 m/s accelerates constantly and reaches a final velocity of 20 m/s in 8 s. Calculate:

• The acceleration of the car
• The displacement if the car moves linearly

### Solution 1

First, let's write the clues for convenience. We have

v0 = 4 m/s
v = 20 m/s
t = 8 s
a = ?
Δx = ?

The acceleration a of the car is calculated using the definition of acceleration. We have

a = v - v0/∆t
= 20 m/s - 4 m/s/8s
= 16 m/s/8s
= 2 m/s2

We can use any of the other equations (ii, iii or iv) to find the car's displacement. For example, if we use the equation iv, we obtain

∆x = v0 × ∆t + a × ∆t2/2
= 4 m/s × 8s + 2 m/s2 × (8s)2/2
= 32m + 64m
= 96m

## Vertical Motion. Free Fall as a Special Case of Vertical Motion

As discussed in the previous tutorial "The Meaning of Acceleration. Constant and Non-Constant Acceleration. Gravitational Acceleration", when objects are at a certain height from the ground, they fall down as the Earth attracts them due to its gravity. We stated that this kind of motion is uniformly accelerated motion when falling down and uniformly decelerated motion when going up. The motion map below helps making this idea clearer.

The interpretation of the motion map is as follows:

"The object is thrown vertically upwards from the ground. It decelerates as the gravity acts against motion. This deceleration occurs until the object reaches its maximum height. Then it stops for a very short instant and afterwards it changes the direction of motion, i.e. it starts falling down. During this process, the object accelerates as the gravity now act in the direction of motion. Therefore, the object reaches the ground moving at a final velocity that is numerically equal to the initial velocity but in the opposite direction."

The gravitational acceleration (denoted by g instead of a to make it distinct from the horizontal acceleration), has two fixed values near the Earth surface (as explained in the previous tutorial). They are +9.81 m/s2 when moving up and -9.81 m/s2 when falling down. Therefore, if writing the symbol of the height h instead of displacement ∆x, the four equations of the motion with constant acceleration discussed in the previous paragraph, become:

v = v0 + g × ∆t

### Equation ii

h = v + v0/2 × ∆t

### Equation iii

v2 - v20 = 2 × g × h

### Equation iv

h = v0 × ∆t + g × ∆t2/2

### Example 2

An object is thrown upwards from the ground at 28 m/s. Calculate

• The maximum height the object can reach
• The instant (s) when the object will be 18 m above the ground.

### Solution 2

The situation is visually described in the figure below

The maximum height hmax is reached when the object after starting moving upwards from the ground (position 0) goes at the highest position (position 1), where it stops for an instant and prepares for turning back at the ground (position 0). Therefore, the velocity in that position is v1 = 0.

Using the equation (III)

v2 - v20 = 2 × g × h

we obtain for the maximum height h after the substitutions (for convenience, let's take the magnitude of gravitational acceleration as 10 m/s2 instead of 9.81 m/s2)

02 - 282 = 2 × (-10) × hmax
-784 = -20 × hmax
hmax = -784/-20
= 39.2 m

(The gravitational acceleration is taken as negative as the object is slowing down when moving upwards).

The object will be twice at 18 m above the ground: one when moving up and the other when falling down. Therefore, since we have to obtain two different results for the time t (or Δt, it is the same thing), we must consider the equation (iv) as it is the only one in which the time t is at power 2 (i.e. is quadratic). Now, the height h is known; it is h = 18m. Thus, we have:

h = v0 × ∆t + g × ∆t2/2
18 = 28 × t + (-10) × t2/2
18 = 28 × t - 5 × t2
5 × t2 - 28t + 18 = 0

As stated before, this is a quadratic equation, i.e. an equation of the form

A × x2 + B × x + C = 0

where A, B and C are constants and x is the variable. In our example, we have A = 5, B = -28, C = 18 and x = t. Therefore, since the discriminant

Δ = B2 - 4 × A × C
= 282 - 4 × 5 × 18
= 784 - 360
= 424

is positive, we have two different roots for this equation which represent the required times. Hence,

t1 = -B - √/2 × A
= -(-28) - √424/2 × 5
= 28 - 20.6/10
= 7.4/10
= 0.74 s

and

t2 = -B + √/2 × A
= -(-28) + √424/2 × 5
= 28 + 20.6/10
= 48.6/10
= 4.86 s

It is obvious the first solution (t1 = 0.74 s) represents the time in which the object is 18 m above the ground when rising up, while the second solution represents the time in which the object is again at 18 m above the ground but this time when it ia falling down. The figure below clarifies this point.

## Free Fall as a Special Case of Vertical Motion

If an object is released from a certain height h, this process is known as "free fall." This is a kind of vertical motion but with some specifics. Thus, in free fall, the objects start falling from rest, i.e. their initial velocity is zero. Also, the gravity (gravitational acceleration g) is taken as positive. In the previous example, the right side of the figure represents a free fall because the object has started falling down from rest after reaching the maximum height. Therefore, we say, "free fall is a special case of the vertical motion."

Given that in free fall we have a vertical motion with v0 = 0, the four equations of the free fall become shorter than those of vertical motion. Thus, we have

v = g × ∆t

h = v/2 × ∆t

v2 = 2 × g × h

h = g × ∆t2/2

### Example 3

An object released from a height h is at h1 = 25 m when its velocity is v1 = 20 m/s. Calculate:

• The initial height h from which the object has fallen down
• The object's velocity v just before touching the ground
• The falling time t

### Solution 3

First of all, it is better to draw a figure as it will help a lot understanding the situation.

First, we have to find the height h0 from the starting point to the position 25 m above the ground. Again we will take g = 10 m/s2 for convenience. Thus, using the equation

v21 = 2 × g × h0

we obtain after rearranging and substituting the values

h0 = v21/2 × g
= 202/2 × 10
= 400/20
= 20m

Therefore, the total height h from which the object has been released, is

h = h0 + h1
= 20m + 25m
= 45m

The final velocity v (the velocity just before touching the ground is calculated using the equation

v2 = 2 × g × h

Substituting the values, we obtain

v2 = 2 × 10 m/s2 × 45m
v2 = 900 m2/s2
v = √900 m2/s2
= 30 m/s

The easiest way to calculate the falling time t is by using the first formula

v = g × t

Thus, rearranging we obtain

t = v/g
= 30 m/s/10 m/s2
= 3s

The following table includes all the formulae discussed in this tutorial.

Remarks!

At first glance, it seems there are too many formulae in this Physics tutorial. However, only five of them are basic: one of the uniform motion and four of the uniformly accelerated (decelerated) motion. The other formulae represent expressions of these five basic ones in various contexts.

Acceleration was taken as negative only when slowing down. Thus, in all examples it was implied the object moves in the positive direction. In fact there is also a situation in which acceleration is takes as negative although the object is speeding up. This occurs when the object is moving towards negative. The negative sign of acceleration here exists because of the direction, not because of the deceleration. However, such situations are not included in this tutorial to avoid confusion.

## Whats next?

Enjoy the "Equations of Motion" physics tutorial? People who liked the "Equations of Motion" tutorial found the following resources useful:

1. Physics tutorial Feedback. Helps other - Leave a rating for this tutorial (see below)
2. Kinematics Revision Notes: Equations of Motion. Print the notes so you can revise the key points covered in the physics tutorial for Equations of Motion
3. Kinematics Practice Questions: Equations of Motion. Test and improve your knowledge of Equations of Motion with example questins and answers
4. Check your calculations for Kinematics questions with our excellent Kinematics calculators which contain full equations and calculations clearly displayed line by line. See the Kinematics Calculators by iCalculator™ below.
5. Continuing learning kinematics - read our next physics tutorial: Position v's Time and Distance v's Time Graph
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