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All Right Let’s Start
• Gonna need to know the speed formula:• Speed = distance (d) / time (t)• 50 meters / 25 seconds = 2 m/s Got to know average speed: Total distance over total time: 50m + 25 m divided by 25 s + 15s + 35 s = 75 m / 75 s = 1 m/s
Then:
• Velocity vs. Speed• Scalars vs. Vectors• Displacement at a 90 degree angle• (a^ +b^)= c^• Distance vs displacement
Acceleration
• ALWAYS LIST THE GIVENS!!!!!!!!!!!!!• Formulas:• Vf- Vi / time • Ex: A car accelerates from 20 m/s to 25 m/s in
2.5 s. What is it’s rate of acceleration?• 25 m/s – 20 m/s / 2.5 s = 2 m/s ^• Remember** Acceleration can be negative
Like so:
• A car slows from 30 m/s to 15 m/s in 5 s. What is the rate of acceleration?
• GIVEN:• Vf = 15 m/s Vi = 30m/s t = 5 s• They want to know rate of acceleration (a)• Vf (15m/s) – Vi (30m/s) / 5 s = - 3m/s^
• Got it?
Now a little trickier
• Displacement: From rest• A car accelerates uniformly from rest to a
velocity of 20 m/s in 7s. How far did it travel in this time?
• We want to know displacement. (Distance)• So we use d= ½ (Vi + Vf)(t) = • Recall: This is when a is constant.
Displacement when already moving
• A car with an initial velocity of 5 m/s accelerates uniformly at a rate of 4 m/s^. Find the displacement after 6 s.
• We are already moving so we use this:• (t^) • d= (5m/s)(6s) + ½ (4m/s^)(6s^) • d= 102 m• Notice Final V doesn’t enter or matter!!!!!
Now for Velocity
• A car with an initial velocity of 4 m/s accelerates uniformly at the rate of 5 m/s^. What is the final velocity after 5 s?
• We want to know velocity• Velocity with constant a is:• Vf= Vi + (a) (t)• Vf= 4 m/s + (5m/s^) ( 5s) = 29 m/s
Now Final Velocity given d
• A car accelerates uniformly form rest at a rate of 4 m/s^. What is the velocity of the car after 65 m?
• Always write the givens. It tells us where to go• Equation:^= • Our Equation : = +(2)(4m/s^)(65m)• Solve Vf = + =22.8 m/s
Now on to FREE FALL
• Remember free fall is interested only in the effect of gravity.
• For our purposes we will use -9.8 m/s^ as our acceleration due to gravity.
• Let’s do distance first:• A ball from rest falls for 4 s. How far did the
ball travel?• d = ½ (t^)(a) (Remember a is every second)
Solve
• Easy: (a)• Then = -78.4 m • Now how about Velocity given distance?• A ball falls from rest 30 m above the floor.
How fast is it moving when it strikes the ground? B. How long until it strikes the ground?
We use the velocity equation
• Free Fall Velocity Equation• Vf^ = Vi^ + (2) (a) ( d)• Vf= 30m)• Vf =- 24.2 m/s or 24.2 m/s down
How about Time?
• Same problem• Solve for time (t)• Equation for free fall time.• Vf= Vi + (a) ( t)• Isolate by 1.subtracting Vi from both sides.• Vf-Vi= (a)(t)• Then dividing both side by a : Vf-Vi / a
Now for Horizontally launched projectiles
Treat X and Y separately Let’s consider perfectly horizontal at a constant velocity. (ALWAYS LIST GIVENS)• A ball is thrown horizontally off a cliff with a
height of 22m at a horizontal velocity of 2.0 m/s. How long before the ball hits the ground?
• First we have to find velocity: = )(-22m)
So
It’s a vertical equation so Vi = 0Vf = • Vf = = 20.77 m/s• Then we plug in time equation
(t) = 2.12 s
How about distance?
• How far from the edge did the ball land?• Horizontal Equation for distance (d ord)• + ½ (a) (t^)• Or 2.0m/s)(2.12s) + ½(0)(2.12^)• Or (2.0ms)(2.12 s) =4.2 m
WOOOOOOW!!!That’s a lot!
• Now for them gosh durn angles!!!!!!!• Remember: Solve x and Y separately!!!!!• We will use Cos and Sin buttons
1st Problem
• Jaynes throws a ball with a velocity of 10m/s at an angle of 30 to the horizon. How far will the ball go?
• First draw a diagram• Then list the trig functions• Then list the givens
Jaynes throws a ball with a velocity of 10 m/s at an angle of 30 degrees above the horizontal. How far will the ball go?
8.83 m
Picture
X Problem (COS)
Y Problem (SIN)
t
v
d
t
a
v
v
d
f
i
vtd
?30
sm /10
sm /10
30
sm /66.8)30cos(10
sm /0.5)30sin(10
Trig
sm /66.8
m0
sm /0.5
sm /0.5
2/8.9 sm
atvv if
t)8.9(0.50.5 t8.90.10
t
8.9
0.10
ts 02.1
s02.1
s02.1
)02.1)(66.8(dmd 83.8
Breanna tosses a Frisbee at 4 m/s at 40 degrees below the horizontal off of the top of a cliff. If the coin lands 4 meters from the cliff, how tall is the cliff?
4m/s
Picture
X Problem Y Problem
t
v
d
t
a
v
v
d
f
i
vtd
?sm /4
40
sm /06.3)40cos(4
sm /57.2)40sin(4
Trig
sm /06.3
sm /57.2
2/8.9 sm
2
2
1attvd i
2)30.1)(8.9(2
1)30.1(57.2 d
md 6.11
s30.1
s30.1
))(06.3(4 t
st
st
30.1
30.106.3/4
m4
m4
40
4 m/s
Caleb fires a gun at 300 m/s. What is its maximum range? (What’s the farthest that it can shoot?)
Picture
X Problem Y Problem
t
v
d
t
a
v
v
d
f
i
vtd
Trig
sm /1.212
sm /1.212
2/8.9 sm
atvv if
t8.91.2121.212
t8.92.424
s29.43
)29.43(1.212dmd 9181
?45
sm /300
sm /300
45
sm /1.212)45cos(300
sm /1.212)45sin(300
m0
sm /1.212st 29.43
s29.43
9181m
Now to practice
• 1. A car driven by Amanda accelerates from rest to 20 m/s in 4 s. What is the rate of acceleration?
• 5 m/s^
• 2. A car driven by Mykayla slows from 40 m/s to 20 m/s in 5 s. What is the rate of acceleration?
• -4 m/s^
More
• 3.A car driven by Josh jumping a red light accelerates uniformly from rest to 25 m/s in 5 s. How far did it go in this time?
• 87.5 m
• 4. A car driven by Ryann accelerates uniformly accelerates from 20 m/s at a rate of 5m/s^. Find the displacement after 5 s.
• 162.5m
And more
• 5. A car driven by Timothy with an initial velocity of 10 m/s accelerates uniformly at 3 m/s^. What is the velocity after 6 s?
• 28m/s• 6. A car driven by Jesse accelerates uniformly
from rest at a rate of 5 m/s^. What is the velocity after 70m?
• 26.46 m/s
Keep going
• 7. A ball is dropped by Ian for 7 s. How far did the ball travel?• 137.2 m
• 8. A ball dropped by Elizabeth falls from rest 4 m to the floor. A. How fast was it going when it hit the floor? B. How long did it take to hit the floor?
• A.- 8.85 m/s• B.90 s
•
And going
• 9. A ball is thrown from a cliff by Media with a height of 20 m at a velocity of 5 m/s . How long before it hits the ground?
• 2.02 s
• 10. How far from the edge did the ball land?• 10.1m
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