1.  Some plants have a hard waxy coating on their leaves that helps prevent water loss. In which environment do these plants most likely grow?  Why is it beneficial that these plants have the wax coating?

4.  It measures the rate of change in velocity (speeding up, slowing down, or changing direction)  Ex: a bullet train must stop to load and unload passengers then take off again; the change in its velocity is called acceleration  Acceleration is also defined as the quantity that describes the rate of change of velocity in a given time interval

5.  a avg = ∆ v/ ∆ t = (v f –v i )/(t f - t i )  Acceleration is measured in m/s 2

6.  A shuttle bus slows to a stop with an average acceleration of –1.8m/s 2. How long does it take the bus to slow from 9m/s to 0m/s?

7.  Known  a avg = -1.8m/s 2  v i = 9m/s  v f = 0m/s  a avg = ∆ v/ ∆ t = (v f – v i )/(t f - t i )  Unknown  ? ∆ t

8.  ∆ t = ∆ v/ a avg  ∆ t = (-9m/s)/(-1.8m/s 2 )  ∆ t = 5.0s  Reminder: acceleration has both a direction and magnitude; direction is determined by the sign  Practice problems with acceleration

9.  A man walks to and from work everday. On the way to work, which is 5 miles away, it takes him 0.25hr. On the way home it takes him 0.5hr.  What is his displacement?  What is his average velocity?

10.  A man sprints to work one day. On the way to work, which is 3 miles away, it takes him 0.5hr. If he started from rest and accelerated to a top speed of 5 mi/hr, what was his acceleration?  What is his displacement?

11.  During constant acceleration, the object speeds up by the same amount every second  Because of this constant speed up, the distance the object travels in each time interval is equal to the distance it traveled in the previous time interval plus a constant distance

12.  Displacement depends on acceleration, initial velocity, and time  So for displacement undergoing constant acceleration, the formula changes to

13.  ∆ x = ½(v i + v f ) ∆ t

14.  A race car reaches a speed of 42m/s. It then begins to have a uniform negative acceleration, using its parachute and braking system, and comes to rest 5s later. Find how far the car moved while stopping.

15.  Known  v i = 42m/s  v f = 0m/s  ∆ t = 5s  ∆ x = ½(v i + v f ) ∆ t  Unknown  ? ∆ x

16.  ∆ x = ½(0m/s + 42m/s)(5s)  ∆ x = 105m

17.  Who is the man in the pic?  What is one contribution  he made to physics?

18.  v f = v i +a ∆ t  This leads to an additional equation for displacement with constant acceleration  ∆ x = v i ∆ t + ½a ∆ t 2

19.  If you know  v f, v i, ∆ t  v i, a, ∆ t  Then use  ∆ x = ½(v i + v f ) ∆ t  ∆ x = v i ∆ t + ½a ∆ t 2

20.  A plane starting at rest at one end of a runway undergoes a uniform acceleration of 4.8m/s 2 for 15s before takeoff. What is its speed at takeoff? How long must the runway be for the plane to be able to take off?

21.  Known  a = 4.8m/s 2  ∆ t = 15s  v i = 0m/s  Looking at our formulas, we can solve for x or v f first. We will start with v f.  Unknown  ?v f, ∆ x

22.  v f = v i +a ∆ t  v f = 0m/s + (4.8m/s 2 )(15s)  v f = 72m/s  Now we know v f, v i, a, and ∆ t so we can use either formula to find ∆ x.

23.  ∆ x = v i ∆ t + ½a ∆ t 2  ∆ x = (0m/s)(15s) + ½(4.8m/s 2 )(15s) 2  ∆ x = 540m

24.  We can rearrange our formulas to find time and get  And one more equation to find final velocity after any displacement

25.  A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500m/s 2. What is the velocity of the stroller after it has traveled 4.75m?

26.  Known  v i = 0m/s  a = 0.500m/s 2  ∆ x = 4.75m  Unknown  ?v f

27.  Take square root to find v f  v f = +2.18m/s  All of the equations are on p.58 of your textbook

28.  Practice 2B p.49  Practice 2C p.53  Practice 2D p.55  Practice 2E p.58  Section Review p.59 #1-3