Displacement with constant acceleration How far, how fast?

Review of v avg We have seen a way to calculate v avg already: v avg = Δx / t Let’s look at the Old School way to find an average….

Finding an average… If you wanted to find the average of these numbers: 5,3,7 and 5 You would just add them up and divide by 4 Mmmk.

Finding an average… You can do the same thing with velocities Suppose a car accelerates from 0m/s for 30 sec at 1m/s 2. That’s gonna take forever to do!

Finding an average…. OR… There’s a much easier way: If acceleration is constant (it always will be for us.) average the first and last velocities and you get the same answer. Oh, good!

Two equations We now have two ways to find v avg : Now let’s put them together… Uhhhh…

One equation 1 st : …

One equation 1 st : Now, substitute in 2 nd : …

One equation 1 st : Now, substitute in 2 nd : New Equation: Ok, got it!

Example A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system, and comes to rest 5.5s later. Find the distance that the car travels during braking.

Example A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system, and comes to rest 5.5s later. Find the distance that the car travels during braking. Givens: v i =42 m/st= 5.5 s v f =0 m/sΔx= ?

Example A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system, and comes to rest 5.5s later. Find the distance that the car travels during braking. Givens: v i =42 m/st= 5.5 s v f =0 m/sΔx= ? Formula:

Example A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system, and comes to rest 5.5s later. Find the distance that the car travels during braking. Givens: v i =42 m/st= 5.5 s v f =0 m/sΔx= ? Formula:

Practice For classwork, complete Practice C, Pg. 53