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Final Exam Review
Final Velocity
Free Fall
2D Motion
Final Velocity
• When solving for final velocity use:
• Formula: vf2 = vi
2 + 2aΔx
Example
• A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/s2. What is the velocity of the stroller after it has traveled 4.75m?
Example
• A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/s2. What is the velocity of the stroller after it has traveled 4.75m?
• Givens a = 0.500 m/s2 vi = 0 m/s
• Δx = 4.75 m
Example
• A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/s2. What is the velocity of the stroller after it has traveled 4.75m?
• Givens a = 0.500 m/s2 vi = 0 m/s
• Δx = 4.75 m
• vf2 = vi
2 + 2aΔx
• vf2 = (0m/s)2 + 2(0.500 m/s2)(4.75m)
• vf2 = 4.75 → vf = 2.18 m/s
Practice
• Practice E- Pg. 58
• #3-4 ONLY
• Hint: For questions that require solving for more than vf, see which formula we have already used that will help.
Free Fall
• In free fall problems, a = 9.8m/s2 every time.
• Formulas: vf = vi + at
• vf2 = vi
2 + 2aΔy
• Motion is along the y axis.
x
y
Example
• A pebble is dropped down a well and hits the water 1.5s later. Using the equations for motion with constant acceleration, determine the distance from the edge of the well to the water’s surface.
Example
• A pebble is dropped down a well and hits the water 1.5s later. Using the equations for motion with constant acceleration, determine the distance from the edge of the well to the water’s surface.
• Givens t = 1.5s a = 9.8 m/s2 vi = 0m/s
• Unknowns Δy = ? vf = ? (need to find vf before Δy)
Example
• A pebble is dropped down a well and hits the water 1.5s later. Using the equations for motion with constant acceleration, determine the distance from the edge of the well to the water’s surface.
• Givens t = 1.5s a = 9.8 m/s2 vi = 0m/s
• Unknowns Δy = ? vf = ? (need to find vf before Δy)
• Step 1: vf = vi + at → vf = (0m/s) + (9.8m/s2)(1.5s) = 14.7m/s
Example
• A pebble is dropped down a well and hits the water 1.5s later. Using the equations for motion with constant acceleration, determine the distance from the edge of the well to the water’s surface.
• Givens t = 1.5s a = 9.8 m/s2 vi = 0m/s
• Unknowns Δy = ? vf = ? (need to find vf before Δy)
• Step 1: vf = vi + at → vf = (0m/s) + (9.8m/s2)(1.5s) = 14.7m/s
• Step 2: vf2 = vi
2 + 2aΔy → (14.7)2 = (0m/s)2 + 2(9.8m/s2)Δy
• Δy = 11 m
Practice
• Practice F- Pg 64
• FIRST TWO ONLY
2D Motion
• Key Point: Draw a diagram
• Key Point: Only consider one direction at a time (either “x” or “y”)
• Key Point: Step one will almost always be to solve for time.
Example
• The Royal Gorge Bridge in Colorado rises 321m above the Arkansas River. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock’s horizontal displacement is 45.0m. Find the speed at which the rock was kicked.
Example
• The Royal Gorge Bridge in Colorado rises 321m above the Arkansas River. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock’s horizontal displacement is 45.0m. Find the speed at which the rock was kicked.
• Givens
• x direction y direction
• Δx = 45m Δy = -321m
• vi=? a = -9.8 m/s2
• t = ? Time will always be the same t = ?
Example- Page 98 (Diagram)
• Givens
• x direction y direction
• Δx = 45m Δy = -321m
• vi=? a = -9.8 m/s2
• t = ? Time will always be the same t = ?
• Step 1: Solve for time.
• Δy = vit + ½at2 → -321m = (0m/s)(t) + ½(-9.8m/s2)t2 → t2 = 65.5
• t = 8.09 s
Example- Page 98 (Diagram)
• Givens
• x direction y direction
• Δx = 45m Δy = -321m
• vi=? a = -9.8 m/s2
• t = 8.09 s t = 8.09 s
• Step 1: Solve for time.
• Δy = vit + ½at2 → -321m = (0m/s)(t) + ½(-9.8m/s2)t2 → t2 = 65.5
• t = 8.09 s
• Step 2: Solve for vi in the x-direction.
• Δx = vit → 45m = vi(8.09s)
• vi = 5.56 m/s
Practice
• Practice D- Pg 99
• FIRST TWO
