Kinematics is the branch of physics that addresses the motion of objects without reference to the forces that cause the motion

Kinematics are very important, as they are the foundation of many of the other units that you will study in physics. To understand the basics of kinematics, you must have a solid grasp on each of the kinematic variables: d, t, vi, v, a. The variable d refers to displacement, which is the distance from the initial point to the resulting point. It is important to not confuse this with distance, which is the total distance that an object has covered. The variable t refers to time passed. The variable v0 refers to the initial velocity, which is the object’s starting velocity. The variable v refers to the final velocity. Lastly, the variable a refers to the acceleration of the object, which is the change in the object’s velocity over the change in time.

It is also important to understand the relationship between displacement, velocity, and acceleration. Velocity is the slope of a displacement-time graph, and acceleration is the slope of a velocity-time graph. You can also use the area under graphs to help you find information (the product of the axes is what the area under a graph will inform you about).

There are one-dimensional kinematics and two-dimensional kinematics. The “Big Five” equations are what we can use to solve for missing variables when address kinematic problems, both in one-dimension and two-dimensions.

Example Problem #1

If an object moves at a velocity of 20 m/s E (positive direction) and slows down to a speed of 5 m/s W over a time of 4 seconds, what is the object’s acceleration?

Answer: -6.25 m/s^2

Explanation: Since we were given the quantities vi, v, and t, we can use Kinematic Equation (KE) #2 to solve for acceleration.  If we plug in +20 m/s for V0, -5 m/s for V, and 4s for t, we can solve for the acceleration and get -6.25 m/s^2.

2-D Kinematics; what's different?

For two-dimensional kinematics, things get a little bit trickier.  Now, we are going to have to analyze projectile motion by using trigonometry and breaking each of the kinematic variables into their x and y components.  Then, we will use each component to help us solve for the missing variable.  Some important notes for 2-D kinematics are that there is zero horizontal acceleration, projectiles follow parabolic paths, and that vertical acceleration of a projectile is always a constant -9.8 m/s^2.

Example Problem #2

If a baseball is launched from the ground at an initial speed of 15 m/s at an angle of 42 degrees above the horizontal, how long will the ball be in the air, and how far will it travel?

Answer: t = 2.048s, dx = 22.829m

Explanation: Break into components

Y: Given: viy, ay, dy → KE #3 → 0 = 15sin(42)t + ½(-9.8)t^2 → t = 2.048s 

X: Now we know t = 2.048s, we can use the fact that dx = vx(t) to solve for dx → dx = (15cos(42))(2.048) = 22.829m