Displacement with constant acceleration

Displacement with constant acceleration

How far, how fast?

Review of vavg

• We have seen a way to calculate vavg already:

vavg = 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 5You would just add them up and divide by 4

54

20

4

)5735(

.#.

)...(

numbersof

numberstheofsumMmmk.

Finding an average…

• You can do the same thing with velocities• Suppose a car accelerates from 0m/s for 30

sec at 1m/s2.

That’s gonna take forever to do!

???30

/30.../2/1/0

smsmsmsmvavg

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!

2fi

avg

vvv

Two equations

• We now have two ways to find vavg :

• Now let’s put them together…

2fi

avg

avg

vvv

andt

xv

Uhhhh…

One equation

• 1st: tvxt

xv avgavg *

…

One equation

• 1st:

• Now, substitute in

• 2nd:

tvxt

xv avgavg *

2fi

avg

vvv

tvv

x fi )2

(

…

One equation

• 1st:

• Now, substitute in

• 2nd:

• New Equation:

tvxt

xv avgavg *

2fi

avg

vvv

tvv

x fi )2

(

tvvx fi )(2

1

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: vi=42 m/s t= 5.5 s vf=0 m/s Δx= ?

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: vi=42 m/s t= 5.5 s vf=0 m/s Δx= ?

• Formula: tvvx if )(2

1

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: vi=42 m/s t= 5.5 s vf=0 m/s Δx= ?

• Formula:

mx

ssmx

ssmsmx

tvvx if

5.115

5.5)/42(2

1

5.5)/42/0(2

1

)(2

1

Practice

• For classwork, complete Practice C, Pg. 53