What does each letter/variable mean?
S = Distance/Displacement (Meters
U = Initial Velocity (Meters per Second)
V = Final Velocity (Meters per Second)
A = Acceleration (Meters per Second per Second)
T = Time (Seconds)
Suvat equations are the basics of mechanics and a key part of A-Level maths. They are some of the first equations you will learn and can span from solving simple problems to some that are very complex. The equations link Velocity, time, displacement and acceleration. However for these equations acceleration has to be constant (Otherwise you need differential equations). Most other parts of mechanics are built at least partially on SUVAT equations and because of this it is important that you just don't memorize the equations but actually understand how they work. In this article I will explain how these equations are arrived, not only mathematically but also logically.
Each of the equations miss out one variable, thus out of the 5 equations you need to have any 3 of the variables and you can work out the fourth.
Layers of Time
This first part isn't strictly SUVAT, however its another piece of theory that is very important to understand.
All of the parts of the equations are related logically, not just as part of the equations. They are all linked by either dividing or timesing by T (Time)
Displacement / Time = Velocity
Velocity / Time = Acceleration
This can then be repeated in reverse by multipling by Time to go back up from acceleration to Velocity and then back to displacement. This relationship is crucial to understand.
S = ((U+V)/2 ) * T
This first equation is the simplest of the SUVAT equations. It uses the the two velocities and time to work out the displacement.
The (U+V)/2 gives you the mean (average) velocity over the period of acceleration. Then multiplying the velocity by Time gives you displacement (As explained by the section of this article to the right.).
V = U+AT
This equation is also one of the simpler SUVAT equations. This time you rely on having the initial velocity, the rate of acceleration and the time, with these you can calculate the final velocity.
The only difficult part of this equation to understand is the reason for the AT. Acceleration multiplied by Time gives you the change in velocity over the time period in question (Refer to my Layer of T section for more explanation.).
If you are starting from stationary then the end velocity is only the change in velocity. However if you start at a speed then you have to add this on, hence the U+. Overall this completes the whole V=U+AT equation.
S=UT+(1/2)AT^2 and S=VT-(1/2)AT^2
These two equations are more complex to understand, however are also both very similar. They require you to have one of the two velocities, acceleration and time to be able to calculate the displacement.
The easiest way to think about understanding these equations is to see a graph. I have drawn one below.
The area under the graph (as shaded) is Displacement (Velocity x Time). You can calculate this area by splitting up the area.
S = UT +(1/2)AT^2 The bottom section is simply Initial Velocity x Time (UT) . The top triangle is given by 1/2 x T x (V-U). For this equation however we don't want both V and U to get change in velocity, luckily however as seen in the equation V=U+AT change in velocity can be given by Acceleration x Time. This means the area of the triangle can be written as 1/2 x AT x T. This gives the (1/2)AT^2. So by combining both areas you get the full equation.
S= VT -(1/2)AT^2 The process for this equation is very similar, however you take the overall area using VT and then minus the same triangle to get the shaded area.
V^2 = U^2 + 2AS
This equation is by far the most complex to derive logically, however is easier to explain mathematically using combinations of the above equations. This equation can be used to give final velocity when you have the initial velocity, the acceleration and the distance traveled.
If you rearrange V=U+AT you can get T= (V-U)/A (Which makes sense if you think of the units (Meters per second divided by Meters per second per second, this gives just seconds, and thus the time.)
Substitute this into S=((V+U)/2)*T and you get S = ((U+v/2))*((V-U)/A). Times the left hand side by 2A to get rid of the denominator. You then can use the box method to multiply (U+V)(V-U), this gives you UV+V^2-U^2-UV, which simplifies to V^2-U^2. So 2AS = V^2 -u^2 , which then can easily be rearranged into the equation you want, V^2 = U^2 +2AS.
I'm sorry that this isn't a logical derivation however it was very hard to come up with one that was any simpler than explaining the equation.
All of the equations above can be rearranged to give any of the other of the variables as subjects. To do this you simply follow the normal rules of rearranging equations.
julie on November 02, 2016:
Thank you at last a clear explanation!! fab