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Physics Lesson 3.8.2 - Equations of the Uniformly Accelerated (Decelerated) Motion

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Welcome to our Physics lesson on Equations of the Uniformly Accelerated (Decelerated) Motion, this is the second lesson of our suite of physics lessons covering the topic of Equations of Motion, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

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

Equation i

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

You have reach the end of Physics lesson 3.8.2 Equations of the Uniformly Accelerated (Decelerated) Motion. There are 4 lessons in this physics tutorial covering Equations of Motion, you can access all the lessons from this tutorial below.

More Equations of Motion Lessons and Learning Resources

Kinematics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
3.8Equations of Motion
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
3.8.1Equations of the Uniform Motion
3.8.2Equations of the Uniformly Accelerated (Decelerated) Motion
3.8.3Vertical Motion. Free Fall as a Special Case of Vertical Motion
3.8.4Free Fall as a Special Case of Vertical Motion

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