1. WE CAN ONLY USE THESE IN ONE DIRECTION AT A TIME (only X or only Y not both at same time)

2. Equation 1 Where V f = Final Velocity V 0 = Initial Velocity a = Acceleration t = Time

3. Equation 2 Where V f = Final Velocity V 0 = Initial Velocity a = Acceleration ∆x = change in distance

4. Equation 3 Where V f = Final Velocity V 0 = Initial Velocity t = time ∆x = change in distance

5. Equation 4 Where t = time V 0 = Initial Velocity a = Acceleration ∆x = change in distance

6. Steps to Solve a kinematic Problem 1. Draw a picture of the problem 2. Identify what we know and label them with the proper variables. 3. Check to see if there are any hidden values not given in number form (Ex: starts from rest) 4. Identify which variable the question is asking you to solve for. 5. Choose the equation to use based on what we have and what we want. 6. Plug and Chug 7. Check to see if your answer seems reasonable

7. Example Step # 1 Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s 2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) Step # 1

8. Example Steps 2-4 Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s 2, then determine the displacement of the car during the skidding process. Step 2 and 3 Step 4 GivenNeed v i = +30.0 m/s ∆x v f = 0 m/s a = - 8.00 m/s 2

9. Example Steps 5 -7 GivenNeed v i = +30.0 m/s ∆x v f = 0 m/s a = - 8.00 m/s 2 Step 5 Equation v f 2 = v i 2 + 2 a ∆x Step 6 0 = 30 2 +2(-8) ∆x  -900=-16∆x  56.25 =∆x