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SUVAT Equations
SUVAT Equations Revision
SUVAT Equations
There are 5 SUVAT equations which relate 5 different variables of motion.
These 5 variables are:
- \textcolor{red}{s} = Displacement
- \textcolor{red}{u} = Initial velocity
- \textcolor{red}{v} = Final velocity
- \textcolor{red}{a} = Acceleration
- \textcolor{red}{t} = Time taken
Questions will usually give you three variables, so you will have to figure out which equation to use.
Note: These equations can only be used when the acceleration is constant.
Key Equations
- \textcolor{red}{v} = u + \textcolor{blue}{a}\textcolor{purple}{t}
- \textcolor{limegreen}{s} = u\textcolor{purple}{t} + \dfrac{1}{2}\textcolor{blue}{a}\textcolor{purple}{t}^2
- \textcolor{limegreen}{s} = \dfrac{1}{2}(u + \textcolor{red}{v})\textcolor{purple}{t}
- \textcolor{red}v^2 = u^2 + 2\textcolor{blue}{a}\textcolor{limegreen}{s}
- \textcolor{limegreen}{s} = \textcolor{red}{v}\textcolor{purple}{t} - \dfrac{1}{2}\textcolor{blue}{a}\textcolor{purple}{t}^2
Paired Equations of Motion
We can use these equations to compare the motion of two (or more) separate particles.
For example, say we have two cars driving on the motorway. Car 1 is in the slow lane, at a constant rate of 50\text{ mph}. Car 2 is overtaking from \textcolor{limegreen}{160}\text{ m} behind, with an initial velocity of 60\text{ mph} and accelerates at a rate of \textcolor{blue}{+2}\text{ mph per second}.
Take \textcolor{limegreen}{160}\text{ m} = \textcolor{limegreen}{0.1}\text{ miles}.
Show that it will take approximately \textcolor{purple}{36}\text{ seconds} for Car 2 to overtake Car 1.
Car 1: u_1 = 50, \textcolor{blue}{a_1 = 0}, \textcolor{limegreen}{s_1 = x}\\
As there is no acceleration, we can use the equation \textcolor{limegreen}{s} = u\textcolor{purple}{t}\\
\textcolor{limegreen}{x} = 50\textcolor{purple}{t}\\
Car 2: u_2 = 60, \textcolor{blue}{a_2 = 2}, \textcolor{limegreen}{s_2 = x + 0.1}\\
As we have the displacement, initial velocity, time taken and acceleration, we can use this SUVAT equation:
\textcolor{limegreen}{s} = u\textcolor{purple}{t} + \dfrac{1}{2}\textcolor{blue}{a}\textcolor{purple}{t}^2\\ \textcolor{limegreen}{x + 0.1} = 60\textcolor{purple}{t} + \textcolor{purple}{t}^2\\
Solve these 2 equations simultaneously:
\textcolor{limegreen}{x} = 50\textcolor{purple}{t}\\ \textcolor{limegreen}{x + 0.1} = 60\textcolor{purple}{t} + \textcolor{purple}{t}^2\\50\textcolor{purple}{t} + 0.1 = 60\textcolor{purple}{t} + \textcolor{purple}{t}^2\\ \textcolor{purple}{t}^2 + 10\textcolor{purple}{t} - 0.1 = 0\\ \textcolor{purple}{t} = \dfrac{-10 ± \sqrt{100 + 0.4}}{2} = 0.00999...\text{ hrs and } -10.00999... \text{ hrs}
Since time cannot be negative, the answer is
0.00999...\text{ hrs} = \textcolor{purple}{35.96}\text{ seconds} ≈ \textcolor{purple}{36}\text{ seconds}
Note:
If a question states that an object is accelerating under gravity, take a to be g = 9.8\text{ ms}^{-2}, unless stated otherwise.
Example 1: Horizontal Motion
A car is driving at 20\text{ ms}^{-1}. After \textcolor{purple}{10}\text{ seconds}, its speed is \textcolor{red}{50}\text{ ms}^{-1}.
How far does it travel in this time? What is its rate of acceleration?
[2 marks]
To work out the displacement, we need an equation using this and the known variables: displacement, initial velocity, time taken, and final velocity:
\textcolor{limegreen}{s} = \dfrac{1}{2}(u + \textcolor{red}{v})\textcolor{purple}{t}\\\textcolor{limegreen}{s} = \dfrac{1}{2}(20 + \textcolor{red}{50}) \times \textcolor{purple}{10} = \textcolor{limegreen}{350}\text{ m}\\
And then acceleration is calculated by dividing the change in velocity by the time taken:
\textcolor{blue}{a} = \dfrac{(\textcolor{red}{50} - 20)}{\textcolor{purple}{10}} = \textcolor{blue}{3}\text{ ms}^{-2}
Example 2: Vertical Motion
Danielle jumps from a diving board \textcolor{limegreen}{10}\text{ m} above a pool. If her initial velocity (upwards) is 1\text{ ms}^{-1}, how long does it take her to reach the water? What is her velocity at this point?
Assume g = \textcolor{blue}{10}\text{ ms}^{-2}.
[4 marks]
To find out time taken to reach the water, we need to use a SUVAT equation that uses time taken, displacement, initial velocity, and acceleration (as we can use gravity as acceleration):
\textcolor{limegreen}{s} = u\textcolor{purple}{t} + \dfrac{1}{2}\textcolor{blue}{a}\textcolor{purple}{t}^2\\ \textcolor{limegreen}{10} = -\textcolor{purple}{t} + 5\textcolor{purple}{t}^2\\ 5\textcolor{purple}{t}^2 - \textcolor{purple}{t} - \textcolor{limegreen}{10} = 0\\ \textcolor{purple}{t} = \textcolor{purple}{1.52}\text{ s (to } 2 \text{ dp)}\\
And to work out her final velocity, we can use:
\textcolor{red}v^2 = u^2 + 2\textcolor{blue}{a}\textcolor{limegreen}{s}\\ \textcolor{red}{v}^2 = (-1)^{-2} + (2 \times \textcolor{blue}{10} \times \textcolor{limegreen}{10})\\ \textcolor{red}{v} =\sqrt{201} = \textcolor{red}{14.18}\text{ ms}^{-1}
SUVAT Equations Example Questions
Question 1: Luke drops a ball from a height of 15\text{ m}. What velocity does the ball hit the ground at?
[1 mark]
Take g = 9.8\text{ ms}^{-2}.
From the data provided, we have s = 15\text{ m}, u = 0\text{ ms}^{-1}, and a = g.
We need to find v.
Using the equation v^2 = u^2 + 2as, we have
v^2 = 0^2 + (2 \times 9.8 \times 15)
v^2 = 2 \times 9.8 \times 15 = 294
v = \sqrt{294} = 17.15\text{ ms}^{-1}
Question 2: A plane is coming in to taxi at an airport. It lands parallel to the surface at 480\text{ ms}^{-1}, and takes exactly one minute to come to a complete stop. What is its acceleration?
[2 marks]
We know that u = 480\text{ ms}^{-1}, v = 0\text{ ms}^{-1} and t = 60\text{ s}.
v = u + at gives
0 = 480 + 60a
a = \dfrac{-480}{60} = -8\text{ ms}^{-2}
(A negative acceleration implies an object is slowing down or accelerating in the opposite direction)
Question 3: James, Jeremy and Richard are having a quarter-mile race. From rest, Richard accelerates at a constant rate of 8\text{ mph/s}. Jeremy sets off one second earlier than he should, and accelerates at 7\text{ mph/s}. James is given a 3 \text{ second} head start on the other two, but his acceleration is only 5.5\text{ mph/s}. Who crosses the line first? Who’s last?
[3 marks]
For all three, u = 0\text{ s and }s = 0.25\text{ mi}.
James:
a = 5.5\text{ mph/s} = 0.00152777...\text{ mi/s}\\
t = \sqrt{\dfrac{0.25 \times 2}{0.00152777}} = 18.09\text{ s}
Jeremy:
a = 7\text{ mph/s} = 0.0019444... \text{ mi/s}\\
t = \sqrt{\dfrac{0.25 \times 2}{0.0019444}} = 16.04\text{ s}
Richard:
a = 8\text{ mph/s} = 0.00222... \text{ mi/s}\\
t = \sqrt{\dfrac{0.25 \times 2}{0.00222}} = 15\text{ s}
Removing 3 \text{ seconds} from James and 1 \text{ second} from Jeremy, we have
Richard – 15\text{ s}
Jeremy – 15.04\text{ s}
James – 15.09\text{ s}
