Concept Overview
- Kinematic variables: displacement, distance, velocity, speed, acceleration, and time.
- Kinematic Equations
- Average speed vs average velocity
- Motion in one dimension – cars, balls, and any object moving in one direction
- Motion in two dimensions (projectiles)- object dropped from a moving plane, a bullet fired horizontally, balled kicked at \(45^\circ\), etc.
- Motion Graphs: displacement vs time, velocity vs time, acceleration vs time.
Kinematic Variables
There are five kinematic variables: \( \Delta x, \, v_0,\, v_f,\, a,\, t \)
Velocity \(v \) is split into two variables: Final Velocity \(v_f \) and Initial Velocity \(v_0 \).
Time is measured in seconds and doesn’t have a direction.
Let’s cover misconceptions.
Displacement
Displacement, \( \Delta x \), (a vector quantity) is how far an object has moved from its origin. Distance (a scalar quantity) is the total amount traveled.
Suppose you take a 200 km trip to San Fransisco and back. Your displacement is zero (because you’re back where you started). Your distance, however, is 400 km (200 km to and 200 km back).
Velocity
Velocity (a vector quantity) is how fast displacement changes. Speed is how fast distance changes.
Suppose you run 1 lap around a 100 meter track in 100 seconds. Your speed is 1 m/s. And your velocity is 0 m/s.
Acceleration
Acceleration \(a \) (a vector quantity) is how fast velocity changes.
Think of (1) speeding up or (2) slowing down in your car.
An acceleration can also be caused by (3) a change in direction of velocity.
Time
Time (a scalar quantity) is often overlooked. But it’s the key to solving many projectile motion questions. Why?
Because time is the SAME in the X and Y direction (more on why this matters below).
Kinematic Equations
Couple of things to remember regarding variable notation:
- [katex] \Delta [/katex] = delta. It means “change in,” which literally means “final – initial.”
- [katex] s [/katex] = distance. [katex] \Delta x [/katex]= change in horizontal displacement. [katex]\Delta y [/katex] = change in vertical displacement.
- [katex] v_0 [/katex] = initial velocity. It is pronounced “v knot.”
- Equations apply both horizontally and vertically
- Use subscripts in the equations to keep track of direction. For example, initial velocity in the vertical direction can be written as, “[katex] v_{0y}[/katex].” Displacement in the vertical direction can be written as [katex] \Delta y [/katex].
Memorize
Memorizing the kinematic equations goes a long way. It only takes around 10 minutes.
Once memorized, you solve kinematic problems 30-50% faster than students that haven’t.
The best way to memorize is to write them down on a flashcard and practice 20 kinematic problems.
Here are the equations:
- \(v_f = v_0 +at \)
- \((v_f)^2 = (v_0)^2 + 2a\Delta x \)
- \(\Delta x = v_0t + \frac{1}{2}at^2\)
- \(\Delta x = v_ft – \frac{1}{2}at^2 \)
- \( \Delta x = \frac{1}{2} t (v_f + v_0) \)
Average Speed vs Average Velocity
While we already covered velocity above, this topic seems to confuse students a bit. So here’s a more in depth explanation.
There is big difference between average speed and velocity.
Average speed is distance divided by time: \(speed = \frac{d}{t} \).
Average velocity is displacement divided by time: \(\overline{v} = \frac{\Delta x}{\Delta t}\).
Example: suppose you run 100 meters north, then back 100 m south, in 100 seconds. Using the equations from above we can find that
a) your average speed is = 200 m ÷ 100s = 2 m/s
b) your average velocity is 0m ÷ 100s = 0m/sec. (Remember that displacement is 0, since you end back up at your starting position)
Motion in 1-Dimension
Any object moving in a straight line ( up, down, left, right ) is moving in one direction.
To problem solve follow this simple framework:
- Write down the 3 knowns and 1 unknown
- Pick a kinematic equation that has all the given variables
- Solve for unknown and plug in numbers
Picking the Right Equation
Here’s a hack to figure out what equation to use: Pick the equation that doesn’t contain the variable you don’t need at all.
For example, in the previous 1D question there is: \( \Delta x, \, v_0,\, v_f,\, t \). We don’t care at all about \(a\) Thus you should use the equation: \(\Delta x = \frac{1}{2} t (v_f + v_0) \).
If thats confusing, I made a short YT video going over an example:
Try it out yourself:
Motion in 2-Dimensions
Objects moving in 2-dimensions, are moving both horizontally and vertically. These objects are called projectiles. Solve problems the exact same way as with one-dimensional kinematics, except apply the formulas in both directions.
Helpful Tips to Remember
- Time stays the same in both the x and y directions
- Horizontal speed, distance, or time, can be found using [katex] \Delta x = vt [/katex]
- The vertical acceleration is almost always, acceleration due to gravity ([katex]9.81 \, m/s^2[/katex])
- Unless given, the initial velocity is almost always 0.
- Vertical velocity at the top of the trajectory is always 0, but horizontal velocity remains constant.
- If a projectile is traveling at an initial angle, use trig to split the vector into horizontal and vertical components.
- The final velocity of the object is the SUM of the horizontal and vertical components of velocity
Solving 2d Motion Problems (Example)
Try it out:
Motion Graphs
Most students overlook graphing. It comes up quite a bit on the AP Exam. I’ll show you the best way to understand how these graphs work. First lets try a problem:
From the acceleration-time graph below, find the total change in velocity in the first 3 seconds.
First, let’s take a step back and go back to definitions
- Velocity is the change in displacement per unit of time. “Change in” is another word for slope/derivative! Thus the slope of the displacement vs time graph is velocity!
- Acceleration is the change in velocity per unit of time. Thus the slope of the velocity vs time graph is acceleration.
Look at it like this: \(\Delta x \rightarrow \Delta v \rightarrow a \). To get to the next one just find the slope/derivative!
Now go backwards: \(a \rightarrow \Delta v \rightarrow \Delta x \). To get to the next one just find the opposite of the derivative: the integral (a fancy way of saying area under the curve!)
Are you able to solve the problem above now?
Here’s another one:
The graph shows the acceleration as a function of time for an object that is at rest at time \( t = 0 \) \( \text{s} \). The distance traveled by the object between \( 0 \) and \( 2 \) \( \text{s} \) is most nearlyLet’s keep practicing with the problems at the end of this post!
Practice Questions for Mastery
A plane, 220 meters high, is dropping a supply crate to an island below. It is traveling with a horizontal velocity of 150 m/s. At what horizontal distance must the plane drop the supply crate for it to land on the island? Use \( g = 9.81 \, m/s^2\).
Bonus – Challenge Questions
These are much harder questions, typically seen on an AP Exam.
Even More Free Practice
Try UBQ OR UBQ Quiz Labs where you can find 1000s of questions and quizzes.
Here’s a comprehensive 10 question quiz on projectile motion.
Here’s another quiz on all of kinematics.
Conclusion
That wraps up everything you need to know for kinematics.
So if you are still having trouble answer these questions, try working with a teacher or professional physics mentor to iron out any mis conceptions.
Up next, we have a Speed Review for Linear Forces!
