The SUVAT Equations: Describing Motion
Understanding the SUVAT Variables
Before diving into the equations themselves, it's crucial to understand the building blocks: the five SUVAT variables. Each represents a specific aspect of an object's motion.
| Symbol | Quantity | SI Unit | Description |
|---|---|---|---|
| $ s $ | Displacement | meter ($ m $) | The overall change in position; a vector quantity (has both magnitude and direction). |
| $ u $ | Initial Velocity | meters per second ($ m/s $) | The velocity of the object at the start of the time interval. |
| $ v $ | Final Velocity | meters per second ($ m/s $) | The velocity of the object at the end of the time interval. |
| $ a $ | Acceleration | meters per second squared ($ m/s^2 $) | The rate of change of velocity. It must be constant for the SUVAT equations to be valid. |
| $ t $ | Time | second ($ s $) | The duration of the motion being analyzed. |
It is vital to remember that these are vector quantities. This means their direction matters. In one-dimensional motion, we typically assign one direction as positive and the opposite as negative. For example, if upward is positive, then the acceleration due to gravity ($ g $) is $ -9.8 m/s^2 $.
The Four Core SUVAT Equations
Now, let's meet the four main characters of our story. Each equation relates a different combination of the five SUVAT variables, allowing you to solve for an unknown quantity as long as you know three others.
This equation connects velocity, acceleration, and time. It is derived directly from the definition of acceleration: the change in velocity per unit time.
This equation states that displacement is equal to the average velocity multiplied by time. It only works when acceleration is constant, ensuring the velocity changes uniformly.
This is a very useful equation as it connects displacement directly to initial velocity, acceleration, and time. It tells you how far an object has traveled without needing to know its final velocity.
This equation is special because it links velocity, acceleration, and displacement, completely eliminating time. It's perfect for problems where time is not given or asked for.
A Step-by-Step Problem-Solving Strategy
Solving SUVAT problems can be straightforward if you follow a consistent method. This strategy will help you tackle any problem systematically.
Step 1: Read the problem carefully. Identify the object whose motion is being described and the specific time interval.
Step 2: List the known variables. Write down the values of the three SUVAT variables you are given. Crucially, also note the variable you are being asked to find.
Step 3: Choose the right equation. Select the SUVAT equation that includes the variable you want to find and the three known variables. This is the key step. If you know three variables and need a fourth, one of the four equations will fit.
Step 4: Check and assign signs. Decide on a positive direction (e.g., right is positive, up is positive). Assign positive or negative values to $ s $, $ u $, $ v $, and $ a $ based on this convention. A car braking, for instance, has a negative acceleration if its initial velocity is positive.
Step 5: Substitute and solve. Plug the known values into your chosen equation and solve for the unknown. Be careful with the algebra, especially when dealing with squared terms in Equation 4.
Step 6: Interpret your answer. Don't forget the units! Also, consider if the sign of your answer makes sense based on the direction you defined as positive.
Applying SUVAT to Real-World Scenarios
Let's put the theory into practice with some detailed examples that show the SUVAT equations in action.
Example 1: The Speeding Car
A car is traveling at $ 20 m/s $ when it starts to accelerate uniformly at $ 2 m/s^2 $. What is its velocity after $ 5 $ seconds, and how far does it travel in this time?
Solution:
First, list the knowns: $ u = 20 m/s $, $ a = 2 m/s^2 $, $ t = 5 s $.
We need to find $ v $ and $ s $.
For final velocity, we use Equation 1: $ v = u + at $.
$ v = 20 + (2 \times 5) = 20 + 10 = 30 m/s $.
For displacement, we can use Equation 3: $ s = ut + \frac{1}{2}at^2 $.
$ s = (20 \times 5) + \frac{1}{2} \times 2 \times (5)^2 = 100 + (1 \times 25) = 125 m $.
The car is traveling at $ 30 m/s $ after covering a distance of $ 125 m $.
Example 2: The Emergency Stop
A driver traveling at $ 30 m/s $ slams on the brakes, causing the car to decelerate uniformly at $ -8 m/s^2 $. How far does the car travel before it comes to a complete stop?
Solution:
Knowns: $ u = 30 m/s $, $ v = 0 m/s $ (it stops), $ a = -8 m/s^2 $.
We need to find $ s $.
We don't know $ t $, so we use Equation 4: $ v^2 = u^2 + 2as $.
$ (0)^2 = (30)^2 + 2 \times (-8) \times s $.
$ 0 = 900 - 16s $.
$ 16s = 900 $.
$ s = 900 / 16 = 56.25 m $.
The car's braking distance is $ 56.25 $ meters.
Example 3: The Falling Object
A ball is dropped from rest from the top of a $ 80 $-meter-high cliff. Neglecting air resistance, how long does it take to hit the ground below? (Take $ g = 10 m/s^2 $).
Solution:
We define downward as positive. Knowns: $ s = 80 m $, $ u = 0 m/s $ (dropped from rest), $ a = g = 10 m/s^2 $.
We need to find $ t $.
We don't know $ v $, so we use Equation 3: $ s = ut + \frac{1}{2}at^2 $.
$ 80 = (0 \times t) + \frac{1}{2} \times 10 \times t^2 $.
$ 80 = 5t^2 $.
$ t^2 = 80 / 5 = 16 $.
$ t = \sqrt{16} = 4 s $.
It takes the ball $ 4 $ seconds to hit the ground.
Common Mistakes and Important Questions
Q: Can I use SUVAT equations if the acceleration is not constant?
A: No, absolutely not. The SUVAT equations are derived from the assumption of constant acceleration. If acceleration changes (e.g., a rocket burning fuel and getting lighter, or air resistance affecting a falling object), you cannot use these equations directly. More advanced calculus-based methods are required for non-uniform acceleration.
Q: What is the difference between distance and displacement?
A: Distance is a scalar quantity; it only has magnitude (e.g., "I ran 5 kilometers"). Displacement is a vector quantity; it has both magnitude and direction (e.g., "I am 3 kilometers north of my home"). The SUVAT variable $ s $ stands for displacement. If an object returns to its starting point, its total displacement is zero, even if the distance traveled was large.
Q: I always get the sign wrong for acceleration due to gravity ($ g $). What should I do?
A: This is a very common issue. The value of $ g $ is always approximately $ 9.8 m/s^2 $ downward. The sign depends entirely on your chosen coordinate system.
- If you define upward as positive, then $ a = g = -9.8 m/s^2 $.
- If you define downward as positive, then $ a = g = +9.8 m/s^2 $.
The key is to be consistent. Once you choose a positive direction, stick with it for all vector quantities ($ s $, $ u $, $ v $, $ a $).
Footnote
1 SUVAT: An acronym formed from the standard symbols used in the equations of motion: S = Displacement, U = Initial Velocity, V = Final Velocity, A = Acceleration, T = Time.
2 Acceleration: The rate at which an object's velocity changes with time. It is a vector quantity, measured in meters per second squared ($ m/s^2 $).
3 Displacement: The overall change in an object's position from its starting point. It is a vector quantity, measured in meters ($ m $).
4 Vector: A physical quantity that has both magnitude (size) and direction (e.g., force, velocity, displacement).
5 Scalar: A physical quantity that has only magnitude and no direction (e.g., mass, time, distance).