Kinematics Equations

Its all about Kinematics Equations Kinematic Equations the branch of mechanics concerned with the motion of objects without reference to the forces that cause the motion. Two types of Equations Constant Velocity Equations (acceleration = 0) Constant Acceleration questions (velocity is constantly changing) (a≠0 but is a constant…the same)

Uniform Motion Equations ( 𝒂 =0) Δv= Δx/Δt (y instead of x for vertical) Δvavg=xtotal/Δt total (book uses 𝒗 for avg velocity) Note: constant velocity means vinst =vavg xf=xi+vavgΔt (y for vertical)

Displacement with Constant Acceleration Recall that for an object moving at a constant velocity, displacement is equal to average velocity times the time interval. x=vt For an object that is changing velocity and uniformly accelerating, the average velocity can be written. 𝒗= 𝒗𝒇+𝒗𝒊 𝟐 Therefore: x= 𝑽𝒇+𝑽𝒊 𝟐 t

Velocity with Average Acceleration If an object’s average acceleration during a time interval is known, the change in velocity during that time can be found. The definition of average acceleration ā= 𝜟𝒗 𝜟𝒕 = 𝒗𝒇−𝒗𝒊 𝜟𝒕 Solving for final velocity: 𝒗𝒇=𝐯𝐢+ā 𝜟t

Displacement with Constant Acceleration In cases in which the acceleration is constant, the average acceleration, , is the same as the instantaneous acceleration, a. The equation for final velocity can be rearranged to find the time at which an object with constant acceleration has a given velocity. 𝒗𝒇=𝐯𝐢+𝐚 Δt

Displacement with Constant Acceleration If the vi , a and Δt are known, the displacement can be found by combining…. Vf=vi+aΔt 𝒙= 𝑽𝒇+𝑽𝒊 𝟐 Δt Substituting the equation for vf into the 2nd equation results in:(I did not use the Δ for t but its there!!! Book does this all the time) x=𝒗𝒊𝒕+ 𝟏 𝟐 𝒂𝒕𝟐 or xf = x0 + 𝒗𝒊𝒕+ 𝟏 𝟐 𝒂𝒕𝟐

Displacement with Constant Acceleration Solve this equation vf = vi+ a∆t for time…. ∆t= 𝒗𝒇+𝒗𝒊 𝒂 Substituting the equation for “t” into x= 𝑽𝒇+𝑽𝒊 𝟐 ∆t RESULTS IN: 𝒗 𝒇 𝟐 = 𝒗 𝒊 𝟐 +𝟐𝒂∆x (this is known as the “timeless equation”)

The Fab Four of Kinematics xf=𝐱𝐨+𝐯𝒊∆𝐭+ 𝟏 𝟐 𝐚∆𝐭𝟐 𝒗 𝒇 𝟐 = 𝒗 𝒊 𝟐 +𝟐𝒂∆𝒙 I use x and y depending on horizontal or vertical. Many times the ∆ is missing but its there!!!

Oh so important little table!!!! © 2014 Pearson Education, Inc.

Question 1: Coming to a stop As you drive in your car at 15 m/s (just a bit under 35 mph), you see a child’s ball roll into the street ahead of you. You hit the brakes and stop as quickly as you can. In this case, you come to rest in 1.5 s. How far does your car travel as you brake to a stop?

Question 1: Coming to a stop Draw the pictures (particle motion and V-T) Do not forget the v and a vectors

Question 1 Answer Acceleration: Distance Traveled: © 2015 Pearson Education, Inc.

Question 2: Kinematics of a rocket launch A Saturn V rocket is launched straight up with a constant acceleration of 18 m/s2. After 150 s, how fast is the rocket moving and how far has it traveled? © 2015 Pearson Education, Inc.

Question 2: Kinematics of a rocket launch (cont.) Speed: Distance Traveled: © 2015 Pearson Education, Inc.

Question 3: Calculating the minimum length of a runway A fully loaded Boeing 747 with all engines at full thrust accelerates at 2.6 m/s2. Its minimum takeoff speed is 70 m/s. How much time will the plane take to reach its takeoff speed? What minimum length of runway does the plane require for takeoff? USE YOUR WORKSHEET DRAW THE PICTURE!!!!!

Question 3: Calculating the minimum length of a runway (cont.) Time: © 2015 Pearson Education, Inc.

Question 3: Calculating the minimum length of a runway (cont.) Runway Length: © 2015 Pearson Education, Inc.

QUESTION 4 A position-time graph of a bike moving with constant acceleration is shown on the right. Which statement is correct regarding the displacement of the bike? A. The displacement in equal time interval is constant. B. The displacement in equal time interval progressively increases. C. The displacement in equal time interval progressively decreases. D. The displacement in equal time interval first increases, then after reaching a particular point it decreases.

Example Problem (if We have time): Champion Jumper The African antelope known as a springbok will occasionally jump straight up into the air, a movement known as a pronk. The speed when leaving the ground can be as high as 7.0 m/s. If a springbok leaves the ground at 7.0 m/s: How much time will it take to reach its highest point? How long will it stay in the air? When it returns to earth, how fast will it be moving? Answers: A: 0.71 s B: 1.4 s C: 7.0 m/s Answers: A: 0.71 s B: 1.4 s C: 7.0 m/s