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 = vi2 + 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 = vi2 + 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 = vi2 + 2aΔy Motion is along the y axis. y x

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 = vi2 + 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 Givens x direction y direction Δx = 45m Δy = -321m 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