Explanations 
Flashcards 
StudySmarter IA 
Notes 
Study Plans 
Study Sets 
Exams 
Magazine 
Find a job 
Mobile App 
Jump to a key chapter
s = Displacement – the total displacement of the body since the beginning of the measure at a given point of time.
u = Initial velocity – the velocity of the body at the beginning of the measurement.
v = Final velocity – the velocity of the body at the end of the measurement.
a = Acceleration – the constant acceleration of the object throughout the measurement.
t = Time taken – the time elapsed from the start to the end of the measurement.
The five constant acceleration equations
There are five different constant acceleration equations that are used to connect and solve for the variables above. It is a good idea to learn these equations by heart.
\(v = u + at\)
\(s = \frac{1}{2} (u + v) t\)
\(s = ut + \frac{1}{2}at^2\)
\(s = vt - \frac{1}{2}at^2\)
\(v^2 = u^2 + 2 as\)
Note that each equation has four of the five SUVAT variables. Given any three variables, it would be possible to solve for any of the other two variables.
When can you use the SUVAT equations? The SUVAT equations apply for a body moving in a straight line with constant acceleration.
Deriving the constant acceleration equations
Let us look at how we obtain these equations.
Equation 1: By definition, acceleration is the change in velocity per unit time. The following diagram demonstrates the concept.
A body with initial velocity u accelerates with a constant acceleration to attain a final velocity v after time t.
Let us express the definition mathematically.
\[Acceleration = \frac {change \space in \space velocity}{change \space in \space time} \rightarrow a = \frac {(v - u)}{t}\]
Rearranging the above equation, we get the first equation :
\[v = u + at\]
Equation 2: Remember, we are dealing with constant acceleration here. So the average velocity across the duration of motion is \(\frac {u+v}{2}\). Multiplying the average velocity by time gives the displacement. Therefore,
\(s = \frac{u + v}{2} \cdot t\)
This gives us the second equation,
\(s = \frac{1}{2} (u + v) t\)
Equation 3: To obtain the third equation, directly substitute the value of v from the first equation into the second equation.
\(s = \frac{1}{2} (u + v) t \rightarrow s = \frac{1}{2} (u + u + at) t\)
This gives us the third equation,
\(s = ut + \frac{1}{2}at^2\)
Equation 4: To obtain the fourth equation, first rearrange the first equation and express it in terms of u.
\(u = v – at\)
Substitute this value of u into the third equation,
\(s = ut + \frac{1}{2} at^2 \rightarrow s = (v - at) t + \frac{1}{2}at^2\)
This gives us the fourth equation,
\(s = vt - \frac{1}{2}at^2\)
Equation 5: To obtain the fifth equation, first rearrange the first equation and express it in terms of t.
\(t = \frac{v - u}{a}\)
Substitute this value of t into the second equation,
\(s = \frac{1}{2} (u + v) t \rightarrow s = \frac{(u + v) \cdot (v - u)}{2a} \rightarrow 2as = (v + u) (v – u) \rightarrow 2as = v^2 - u^2\)
This gives us the fifth equation,
\(v^2 = u^2 + 2as\)
Solving problems using constant acceleration equations
Let us look at some examples of problems that can be solved using the constant acceleration equations.
A car with an initial velocity of 8 m/s is accelerating at a rate of 2 m/s². How long will it take to reach a speed of 20 m/s?
Solution 1
Here, v = 20 m/s, u = 8 m/s, a = 2 m/s².
\[v = u + at \rightarrow t = \frac{v - u}{a} \rightarrow t = \frac{20 - 8}{2} = 6 s\]
A car with an initial velocity of 8 m/s is accelerating at a rate of 2 m/s². How long will it take to travel a distance of 200 m?
Solution 2
Here, s = 200 m, u = 8 m/s, a = 2 m/s².
\(s = ut + \frac{1}{2}at^2 \rightarrow 65 = 8t + \frac{1}{2} 2t^2 \rightarrow t^2 + 8t – 65 = 0 \rightarrow t = 5\)
Note: The obtained quadratic equation gives two values, 5 and -13. Time cannot be negative, so we take the positive value as the answer.
A marathon runner decides to accelerate during the last 200 meters of a race. He accelerated at a rate of 0.07 m/s², eventually crossing the finish line at a speed of 8 m/s. What speed was he running at before he decided to accelerate?
Solution 3
Here, s = 200 m, v = 8 m/s, a = 0.07 m/s².
\(v^2 = u^2 + 2 as \rightarrow u^2 = v^2 - 2as = 8 \cdot 8 - 2 \cdot 0.07 \cdot 200 \rightarrow u^2 = 36 \rightarrow u = 6 m/s\)
A cyclist is travelling along a straight road. It accelerates at a constant rate from a velocity of 4 m/s to a velocity of 7.5 m/s² in 40 seconds. Find:
a) the distance she travels in these 40 seconds. b) her acceleration in these 40 seconds.
Solution 4
a) \(s = \frac{1}{2} (u + v) t \rightarrow s = \frac{1}{2} (4 + 7.5) \cdot 40 = 230 m\)
b) \(v = u + at \rightarrow 7.5 = 4 + 40a \rightarrow a = \frac{7.5 - 4}{40} = 0.0875 m/s^2\)
A ball is thrown up with an initial velocity of 39.2 m/s. How long will it take the ball to reach its peak height assuming g = 9.8 m/s²
Solution 5
Here, the acceleration of the ball is -9.8 m/s², as it is the force of gravity that is slowing the ball down.
\(v = u + at \rightarrow 0 = 39.2 - 9.8t \rightarrow t = 4 s\)
Constant Acceleration Equations - Key takeaways
The constant acceleration equations are used to connect five different variables: s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time taken.
The constant acceleration equations are applicable for a body moving in a straight line with constant acceleration.
The constant acceleration equations can be derived starting from the basic definition that acceleration is the change in velocity per unit time.
Each constant acceleration equation contains four of the five SUVAT variables. Given any of the three variables of an equation, it should be possible to solve for the fourth variable as the unknown.
Learn with 0 Constant Acceleration Equations flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
Already have an account? Log in
Frequently Asked Questions about Constant Acceleration Equations
What is the equation for constant acceleration?
The equation for constant acceleration is v = u + at, where u= Initial velocity, v= Final velocity, a= Acceleration, t= Time taken
What is the equation for motion with constant acceleration?
There are 5 commonly used equations for motion with constant acceleration :
1) v = u + at
2) s = ½ (u + v) t
3) s = ut + ½at²
4) s = vt - ½at²
5) v² = u² + 2 as
where s= Displacement, u= Initial velocity, v= Final velocity, a= Acceleration, t= Time taken
Q3: What is constant acceleration?
Acceleration is the change in velocity over time. If the rate of change of velocity of a body remains constant over time, it is known as constant acceleration.
What are examples of constant acceleration?
An example of constant acceleration is a body falling under the force of gravity with no other external force acting on it. In reality, it is very difficult to achieve perfect constant acceleration because there are typically multiple forces acting on any object.
What are the five equations of motion?
There are five commonly used equations for motion with constant acceleration
1) v = u + at
2) s = ½ (u + v) t
3) s = ut + ½at²
4) s = vt - ½at²
5) v² = u² + 2 as
where s= Displacement, u= Initial velocity, v= Final velocity, a= Acceleration, t= Time taken.
About StudySmarter
StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.
Learn more