Acceleration

1. Acceleration • When an object moves, most of the time it’s motion (velocity) will not be constant. • Whenever an object changes it’s velocity (speed with a direction) the object has what is called an acceleration. • An object has an acceleration whenever it… • Speeds up • Slows down • Changes direction • An acceleration is how fast the velocity of an object changes. • Since velocity is a vector, acceleration is also a vector.

2. Instantaneous and Average Acceleration • There are two types of acceleration: • Average acceleration– This is how fast an object changes it’s velocity over a period of time. • Instantaneous acceleration– How fast an object changes it’s velocity at a given point in time. • A car driver speeds up from 0 to 50 MPH in 5 seconds. • This takes place in 5 seconds time so this is an average. • This is an average acceleration • A rock is dropped, and accelerates downward at 9.8 m/s/s when released. • The rock accelerates at 9.8 m/s/s at THE INSTANT it is released. • This is an instantaneous acceleration.

3. The Units for acceleration can be any unit of velocity (speed) divided by a unit of time. Vf -Vi Units for Acc: (Meters/sec)/sec or Meters/sec2 (Miles/Hr.)/sec (Feet/min)/sec (Kilometers/Hr.)/min Acc. Time Calculating Average Acceleration Acceleration is how fast the velocity changes. It is the change in velocity in a given time. Change in Velocity Average Acceleration = ------------------------- Time for the change

4. Finally substitute the numbers into the equation and solve. Next cover up the part of the box you are looking for. Vf -Vi Acc = ------------------- Time The equation you want is: Vf - Vi Acc = ------------------- Time Vf -Vi Vf -Vi Vf -Vi First write out the box Acc. Acc. Acc. Time Time Time 8 m/s – 0 m/s Acc = --------------------- = 16 m/s/s or 16 m/s2 .5 sec Using the acceleration box to find the average acceleration • A Girl starts from rest and accelerates to a running speed of 8 m/s in a time of .5 seconds. What is her average acceleration?

5. Next cover up the part of the box you are looking for. Finally substitute the numbers into the equation and solve. The equation you want is: Vf - Vi Time = ------------------- Acc Vf -Vi Time = ------------------- Acc Vf -Vi Vf -Vi Vf -Vi 14 m/s – 6 m/s 8m/s Time = --------------------- = ----------- = 2 sec 4 m/s2 4 m/s2 First write out the box Acc. Acc. Acc. Time Time Time Using the acceleration box to find the time needed to accelerate • A car traveling at 6 m/s accelerates at a rate of 4m/s2. How much time will it take to reach a speed of 14 m/s?

6. Finally substitute the numbers into the equation and solve. Vf = Acc*Time + Vi Next cover up the part of the box you are looking for. The equation you want is: Vf – Vi = Acc. * Time which becomes Vf = Acc*Time + Vi Vf = -9.8 m/s2 * 5 s + 0 m/s Vf = - 49 m/s ( - means DOWN) Vf -Vi Vf -Vi Vf -Vi First write out the box Acc. Acc. Acc. Time Time Time Using the acceleration box to find the final velocity (speed). • A rock is dropped and falls with a constant acceleration of –9.8 m/s2. What will the rock’s velocity be 5 seconds into the fall?

7. Finally substitute the numbers into the equation and solve. Vi = -(Acc*Time + Vf) Next cover up the part of the box you are looking for. The equation you want is: Vf – Vi = Acc. * Time which becomes - Vi = Acc*Time – Vf Multiply both sides by –1 Vi = -(Acc*Time – Vf) Vi = -(-.5 m/s2 * 5 s + 0 m/s) Vi = + 2.5 m/s2 Vf -Vi Vf -Vi Vf -Vi First write out the box Acc. Acc. Acc. Time Time Time Using the acceleration box to find the Initial velocity (speed). • A ball is rolled across a floor with with an unknown speed, and comes to a complete stop in 5 seconds. If the acceleration of the ball is -.5 m/s2, what was the initial speed of the ball?

8. Picturing Acceleration 0 m .5m Let us say that there is a ball that rolls from rest with a constant acceleration of 2m/s2. What would the ball’s motion look like second by second? 2 m 4.5 m 8 m

9. Time (seconds) Position (meters) Position (m) 0 1 2 3 4 0 .5 2 4.5 8 Time (sec) Graphing the Displacement of Acceleration We can construct a data table and a position versus time graph for the motion. The position verses time graph for accelerated motion will ALWAYS be a parabola. The is because the velocity (the slope) is always changing.

10. Position (m) Time (sec) The Position Equation • The Position equation is similar to the equation for a parabola. Y = AX2 + BX + C • The Y axis is the Pos of the object, and X is the time the object is at that position Pos = At2 + Bt + C • C is a constant (sometimes called the initial condition, A.K.A. y-intercept), we can find it by setting t = 0 sec. Pos = A(0 sec)2 + B(0 sec) + C Pos = C • We now have two positions in the equation so we need to distinguish between the final and initial positions • Y = Posfinal = Posf • C = Posinitial = Posi Posf = At2 + Bt + Posi

11. The Position Equation (cont.) • We still need to find the coefficients A and B from the equation: Posf = At2 + Bt + Posi • The coefficient B is initial velocity of the object Vi • The coefficient A is half of the object’s acceleration (1/2)a • So the position equation is: Posf = (1/2)at2 + (Vi)t + Posi • It is also more commonly written as: Posf = Posi + (Vi)t + (1/2)at2

12. Moving in a positive direction Moving in a negative direction Position (m) Position (m) Time (sec) Time (sec) Position (m) Position (m) Moving in a positive direction Moving in a negative direction Time (sec) Time (sec) If you are speeding up, the parabola becomes more curved. That is because the distance traveled each second increases more and more. Speeding Up, Slowing Down If you are slowing down the parabola becomes less curved. That is because the distance traveled each second becomes less and less.

13. Let’s start with the acceleration equation Vf -Vi Acc = ---------------- Time We can rewrite the acceleration equation Now compare it to the equation for a straight line Vf = acc*time + Vi y = m*x + b They are the same!!! Y = Vf M = slope = acceleration B = intercept = Vi Graphing the Velocity of an Accelerated Object

14. Vel. A positive slope means a positive acceleration. time A negative slope means a negative acceleration. Vel. time This means that the velocity graph for a object moving with a constant acceleration is a straight line. THE SLOPE OF THE LINE IS THE ACCELERATION!!!! Graphing the Velocity of an Accelerated Object

15. t An object can have a negative acceleration and still pick up speed. If the object already has a negative velocity. V t An object can have a positive acceleration and slow down. If the object started with a negative velocity. V Remember +/- means direction Important note: If the velocity line moves away from the time (X) axis the object is speeding up. If the velocity line moves to the time (X) axis the object is slowing down.

16. The Time Independent Equation We now have two equation to describe the motion of an object. Posf = Posi + Vit + (1/2)at2 Vf = Vi + at These are the two most commonly used equations, but they both depend on time. And time is not always something we know. By substituting one equation into another we can derive a 3rd equation that does not depend on time. Rewriting the velocity equation we now have the equation: t = (Vf - Vi)/a Substituting this into the position equation we get the following: Posf = Posi + Vi[(Vf - Vi)/a] + (1/2)a [(Vf - Vi)/a] 2 Which can be simplified to the equation Vf2 = Vi2 + 2a(Posf – Posi) Because it does not depend on time it is called the time independent equation

17. The Big 3 Equations of Motion We now have three equations that can be used to solve most Problems. These are the “Big 3 Equations of Motion” Posf = Posi + Vit + (1/2)at2 Vf = Vi + at Vf2 = Vi2 + 2a(Posf – Posi)

18. Equation Know Want Want – Know – Equation Table This is a problem solving strategy that can be used to organize the provided information in a problem and aid in the selection of an appropriate equation. It is called a want-know- equation table. Posf = Posi + Vit + (1/2)at2 Vf = Vi + at Vf2 = Vi2 + 2a(Posf – Posi) Here place what you are looking for. Posf, Posi, Vi, Vf a, or t Here place the information provided from the problem Posf, Posi, Vi, Vf a, and/or t Here you ALWAYS place the “Big 3 Equations of Motion” The equation that will solve the problem will be easy to spot.

19. Equation Know Want Posf = Posi + Vit + (1/2)at2 Vf = Vi + at Vf2 = Vi2 + 2a(Posf – Posi) Example Problem 1 • A box is lifted from rest to a speed of 20 m/s in 5 seconds. What is the acceleration of the box? Vi = 0 m/s Vf = 20 m/s t =5 sec Posi = 0m a From the table we see that the second equation is the only one that will work. The other two require the final position, which we do not want, nor know. We want the acceleration (a) From the problem we know the following: Remember at rest means the velocity is 0m/s

20. Example Problem 1 (cont) Known: Vi = 0 m/s Vf = 20 m/s t =5 s Posi = 0m. Equation: Vf = Vi + at 20 m/s = 0 m/s + a(5 s) 4 m/s/s = a The acceleration of the box is 4 m/s/s, or 4 m/s2

21. Equation Know Want Posf = Posi + Vit + (1/2)at2 Vf = Vi + at Vf2 = Vi2 + 2a(Posf – Posi) Example Problem 2 • A box is pulled from rest across a rough floor with an acceleration of 3 m/s/s. How far will the bob move in 10 seconds? Vi = 0 m/s t =10 sec Posi = 0m a = 3 m/s/s Posf From the table we see that the first equation is the only one that will work. The other two require the final velocity, which we do not want, nor know. We want the Final position (Posf) From the problem we know the following: Remember at rest means the velocity is 0m/s

22. Example Problem 2 (cont) Knowns: VYi = 0 m/s t =10 sec PosXi = 0m aX.= 3 m/s/s Equation Posf = Posi + Vit + (1/2)at2 Posf = 0 m + (0 m/s)(10 s) + (1/2)(3 m/s/s)(10 s)2 Posf = 0 m + 0 m + (1/2)(3 m/s/s)(100 s2) Posf = (1.5 m/s/s)(100 s2) Posf = 150 m