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Displacement with constant acceleration. How far, how fast?. Review of v avg. We have seen a way to calculate v avg already: v avg = Δ x / 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 - PowerPoint PPT Presentation
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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