Learning Objectives

• We should know what Kinematic Equations do.
• We should know how Kinematic Equations are formulated.
• We should know how to use them to describe motion.

Kinematic Equations

• Equations that allow us to track the position and velocity of an object(s) often with respect to time.
• Integrated from the definitions of acceleration and velocity.
• We assume a constant acceleration.

Calculating Velocity

Consider the definition of velocity in one dimension. \[ \frac{dv}{dt} = a \\ dv = a\space dt \\ \int dv = \int a\space dt \\ \int dv = a\int dt \\ v = v_0 + a\space t \]

Where a is the objects acceleration, v is the speed at the end of the time interval t, \(v_0\) is the initial velocity.

Calculating the Position

Now consider the definition of velocity in 1 dimension. \[ \frac{dx}{dt} = v \\ dx = v\space dt \\ \int dx = \int v\space dt \\ \int dx = \int (v_0 + a\space t) dt \\ \int dx = v_0\int dt+ a\space \int t dt\\ x = x_0 +v_0t +\frac{1}{2}at^2 \]

x is the position of the object at the end of time interval t. \(x_0\) is the initial position.

In Practice

• The above derivations were for 1 dimension.
• 2 or 3 dimensional scenarios are just 2 or 3 1D equations.
• Subscripts help keep track of different dimensions, e.g., \(a_y\) or \(v_z\).
• Trigonometric functions may be needed to separate dimensions.
• For multiple objects, each object has its own set of equations. Again subscripts help, e.g., \(x_1\) or \(v_{2,y}\).

Example

A baseball is hit with initial velocity of 14.1 m/s at a 45 degree angle above the ground. How far from home plate does it travel before it is caught at the same height that it was hit. Treat home plate as \(x_0 = 0\)m.

Difference in y

In this scenario, what is \(y-y_0\)?

1. -1.5 m (height of a person from shoulders to feet)
2. 0 m
3. +1.5 m (height of a person from shoulders to feet)
4. Impossible to tell from given information.

First calculate y for flight time.

\[ v_{y,0} = 14.1*sin(45^o) \\ v_{y,0} = 9.97\space\frac{m}{s}\\ a = -9.8\space\frac{m}{s^2} \\ y = y_0 +v_{y,0}t +\frac{1}{2}at^2 \\ y - y_0 = 9.97t - 4.9t^2 \\ 4.9t^2 = 9.97t \\ t = \frac{9.97\frac{m}{s}}{4.9\frac{m}{s^2}} \\ t = 2.03 s \]

Then calculate x for distance

\[ v_{x,0} = 14.1*sin(45^o) \\ v_{x,0} = 9.97\space\frac{m}{s}\\ a = 0\space\frac{m}{s^2} \\ x = x_0 +v_{x,0}t +\frac{1}{2}at^2 \\ x = 9.97\space\frac{m}{s} 2.03s \\ x = 20.2 m \]

A graph of the system

Note that we get the charateristic parabola of projectile motion in Earth’s gravity.