Final Exam Review Final Velocity Free Fall 2D Motion

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