Unit 3 Notes

General Equation 1: v = at + v 0

General Equation 3 Square both sides of equation 1. Distribute (FOIL). Subtract x 0 from both sides. Multiply both sides by 2a.

v 2 = a 2 t 2 + 2atv 0 +v 0 2 The underscored parts are equal. Substitute 2a∆x into the first equation to get a formula for v in terms of x (with no t in it). General Equation 3: v 2 = 2a∆x +v 0 2 2a∆x=a 2 t 2 +2av 0 t

Velocity is the rate of change in position (slope of the x vs t graph) Acceleration is the rate of change in velocity (slope of the v vs t graph). Example: Construct an x vs t and an a vs t graph for the v vs t graph shown.

The slope of v vs t is negative and does not change, so a is negative and does not change. This should look familiar from unit 2.

Velocity is the slope of x vs t. Velocity starts out as some large, positive value and decreases.

Motion Maps Position and velocity are the same as before. The beginning of the arrow shows the object’s position at the beginning of the time interval, and the end of the arrow shows the object’s position at the end of the time interval. A second arrow next to the motion map shows the direction and magnitude of acceleration.

The length of the black arrow shows the velocity (change in position). The length of the red arrow shows the acceleration (change in velocity).

Kinematic Equations When solving word problems, it is important to identify which quantities are known and which you are solving for. Once you have them, look at which equation has all of those values and no others. Once you have identified the quantities and the formula to be used, solve in three steps.

v = at + v 0 x = ½ at 2 + v 0 t + x 0 v 2 = 2ea∆x +v 0 2 Example: A bike is moving at 1m/s. It accelerates at a rate of 3m/s every second for 2s. What is its final velocity? v 0 =1m/st = 2s a = 3m/s 2 v = ?

v 0 =1m/st = 2s a = 3m/s 2 v = ? The formula with all of these values is v = at + v 0 v = at + v 0 v = (3m/s 2 )(2s) + 1m/s v = 7m/s

Example 2: A ball rolling down a hill starts at rest and accelerates at a rate of 2m/s 2. How far does it travel in 4s? Solution: The object starts at rest, so v 0 =0. We are interested in the change in position, so we can let x 0 be 0. v 0 =0v 0 =0 x = ? t = 4sa = 2m/s 2

v 0 =0v 0 =0 x = ? t = 4sa = 2m/s 2 The relevant equation is x = ½ at 2 + v 0 t + x 0 x = ½ at 2 + v 0 t + x 0 x= ½(2m/s 2 )(4s) 2 +(0)(4s) + 0 x=16m