Constant Acceleration (SUVAT Equations)

The "SUVAT" equations are used to calculate the following:

• Displacement (a distance with a direction)
• Initial velocity (initial speed with a direction)
• Final velocity (final speed with a direction)
• Acceleration (must be constant)
• Time

To do this, we need 3 of the other values on that list. It assumes that (1) there is no drag and whatnot and (2) that acceleration is constant - e.g. from gravity.

This post will explain the different formula, where they come from and give some examples of use (including on the vertical axis).

SUVAT Equations
Firstly, let's establish what the letters mean.
s = displacement
u = initial velocity
v = final velocity
a = acceleration
t = time

From those, we can establish a number of equations.

1. v = u + at
2. s = ((v+u)/2)t
3. s = ut + 0.5at^2
4. s = vt - 0.5at^2
5. v^2 = u^2 + 2as

Where these come from

1. The gradient on a velocity-time graph is where this comes from. a = (v-u)/t is the gradient. You rearrange that to at = v- u and then v = u + at.
2. First we find the average velocity from u and v by (v + u)/2. From there, we just times by the times (following the same principles of distance = speed * time). This gives us ((v + u)/2)t.
3. To do this, we first rearrange the second equation to get v as the subject. This gives you v = 2s/t -u. As both the first and this new equation equal v, one can set them to be equal, giving you 2s/t -u = u + at. This can be rearranged to s = ut + 0.5at^2.
4. This one is very similar to equation 3. Instead, we get u to be the subject. Put them equal to each other, which gives us 2s/t - v = v - at. This can be rearranged to get s = vt - 0.5at^2.
5. Following the same ideas as before, this time we use t. Arrange them both to have t as the subject, put them equals to each other and you get (v - u)/a = 2s/(v + u). From this, v^2 = u^2 + 2as can be gotten.

So why do we go through all these steps to get these equations? Using these equations, we can get any of the five aforementioned values from just three of the others. Equation 1 doesn't need s, 2 doesn't need a, 3 doesn't need v, 4 doesn't need u and 5 doesn't need t.

Examples (non-vertical)
I. A car travels along a straight horizontal road. It accelerates uniformly from rest to a velocity of 12 m/s in 5 seconds. Find how far it travels and how long it takes it.

s = ?
u = 0 m/s
v = 12 m/s
a = 5 m/s^2
t = ?

v^2 = u^2 + 2as
-> 144 = 10s
Therefore, s = 14.4m

v = u + at
-> t = (v - u)/a
-> t = 12/5
Therefore, t = 2.4 seconds

II. A cyclist is travelling at 5 m/s. They break steadily, coming to a rest in 50m. Find the time it takes them to come to a rest and their deceleration.

s = 50 m
u = 5 m/s
v = 0 m/s
a = ?
t = ?

v^2 = u^2 + 2as
-> 0 = 25 + 100a
Therefore, a = -0.75 m/s^2 so deceleration = 0.75 m/s^2.

s = ((v + u)/2)t
-> 50 = ((5)/2)t
-> 2.5t = 50
Therefore, t = 20 seconds.

Examples (vertical)
Note: gravity is the factor here, in this I'll assume it's 10 m/s^2 towards Earth but people can be as specific as they want (9.81 m/s^2 can be used and so on). Also, when something thrown upwards reaches its maximum height, its velocity at that moment will be 0.

I. Neglecting air resistance, how far will a brick starting from rest fall freely in 3 seconds?

s = ?
u = 0
v = X
a = 10
t = 3

s = ut + 0.5at^2
= 0*3 + 0.5*10*9
= 45m

II. A ball is thrown upwards and it reaches its maximum height of 11.25 m. Calculate its initial velocity and the time it takes to reach that height.

s = 11.25
u = ?
v = 0
a = -10
t = ?

v^2 = u^2 + 2as
-> 0 = u^2 + 2 * -10 * 11.25
-> 0 = u^2 - 225
-> u^2 = 225
Therefore, u = 15 m/s

s = v - 0.5at^2
-> 11.25 = -0.5 * -10 * t^2
-> 11.25 = 5 * t^2
-> t^2 = 2.25
Therefore, t = 1.5 seconds