Accelerated Motion

Acceleration Copyright © McGraw-Hill Education 3-1: Acceleration The rate at which an object’s velocity changes. Measured in m/s 2 Vector quantity Constant Acceleration: When velocity of an object changes at a constant rate Acceleration can occur when speed is constant, but the direction of the object is changing.

Acceleration Copyright © McGraw-Hill Education 3-1: Nonuniform Motion Diagrams You can determine the length and direction of an acceleration vector by subtracting two consecutive velocity vectors and dividing by the time interval.

Acceleration Copyright © McGraw-Hill Education 3-1: Direction of Acceleration The direction of both the velocity and acceleration are needed to determine whether an object is speeding up or slowing down. Acceleration in the same direction as velocity: speed increases. Acceleration in the opposite direction of its velocity: speed decreases.

Acceleration Copyright © McGraw-Hill Education 3-1: Velocity-Time Graphs Velocity = y-axis Time = x-axis Slope = acceleration

Acceleration Copyright © McGraw-Hill Education 3-1: Average and Instantaneous Acceleration Average acceleration = change in velocity divided by change in time. Instantaneous acceleration: change in velocity at an instant of time Found by drawing a tangent line on the v-t graph at the point of time you need The slope of this line equals instantaneous acceleration.

Acceleration Copyright © McGraw-Hill Education Calculating Acceleration EVALUATE THE ANSWER The units are correct: Acceleration is measured in m/s 2. Use with Example Problem 1. Problem Use the v-t graph below to answer the following questions. a.What is the object’s average acceleration between 20.0 s and 40.0 s? b.What is the object’s instantaneous acceleration at 30.0 s? Response A SKETCH AND ANALYZE THE PROBLEM List the knowns and unknowns. For part a, the velocity at the needed times can be found on the graph. KNOWNUNKNOWN v 20 = 4.0 m/sā = ? v 40 = 0.0 m/s Δt = 20.0 s SOLVE FOR THE UNKNOWN Use the definition of average acceleration.

Acceleration Copyright © McGraw-Hill Education Calculating Acceleration EVALUATE THE ANSWER The units are correct: Acceleration is measured in m/s 2. Use with Example Problem 1. Problem Use the v-t graph below to answer the following questions. a.What is the object’s average acceleration between 20.0 s and 40.0 s? b.What is the object’s instantaneous acceleration at 30.0 s? Response B SKETCH AND ANALYZE THE PROBLEM List the knowns and unknowns. For part b, draw a line tangent to the curve at the point t = 30.0 s and find the slope of the line. KNOWNUNKNOWN (x 1, y 1 ) = (25.0 s, 4.0 m/s)a = ? (x 2, y 2 ) = (40.0 s, 2.0 m/s) SOLVE FOR THE UNKNOWN Calculate the slope of the tangent line to find the instantaneous acceleration. ā = −0.20 m/s 2

Acceleration Copyright © McGraw-Hill Education EVALUATE THE ANSWER The units are correct: Acceleration is measured in m/s 2. The sign agrees with the drawing. A quick estimate reveals that the magnitude is reasonable: 30/9 = 3.3. Use with Example Problem 2. Problem On a test track, a professional driver preforms a 0 to 60 test. That is, she starts the car at rest and accelerates as quickly as possible to 60 miles per hour (28.6 m/s). If the small sedan she is testing has a 0 to 60 time of 8.5 seconds, what is the car’s average acceleration? Response SKETCH AND ANALYZE THE PROBLEM Sketch the situation. List the knowns and unknowns. KNOWNUNKNOWN v i = 0.0 m/sā = ? v f = 28.6 m/s Δt = 8.5 s SOLVE FOR THE UNKNOWN Use the definition of average acceleration. Calculating Acceleration

The slope of a position-time graph of a car moving with a constant acceleration gets steeper as time goes on. Motion with Constant Acceleration Copyright © McGraw-Hill Education 3-2: Position & Constant Acceleration

Position-time graph of constant acceleration = parabola. The slopes from the position time graph can be used to create a velocity-time graph. Area under v-t graph = object’s displacement. Can’t create a position-time graph from the data, though. Motion with Constant Acceleration Copyright © McGraw-Hill Education Position with Constant Acceleration

If acceleration is constant, average acceleration (ā) = instantaneous acceleration (a). Final velocity equation can be rewritten to find: the time at which an object with constant acceleration has a given velocity. initial velocity If both the v f and the time v f occurred are given. Motion with Constant Acceleration Copyright © McGraw-Hill Education Velocity with Average Acceleration Final Velocity with Average Acceleration

Motion with Constant Acceleration Copyright © McGraw-Hill Education EVALUATE THE ANSWER 6.0 m is a reasonable length for a driveway. The negative sign indicates that the car was backing up. Use with Example Problem 3. Problem The v-t graph below represents the motion of a car backing out of a driveway. What is the car’s displacement at t = 4.0 s? Response ANALYZE THE PROBLEM List the knowns and unknowns. KNOWNUNKNOWN Δt = 4.0 sv = ? Δx = ? SOLVE FOR THE UNKNOWN Use the graph to determine the velocity. v = −1.5 m/s Use the relationship among displacement, velocity, and time. Δx= vΔt = (−1.5 m/s)(4.0 s) = −6.0 m Velocity with Average Acceleration

The graph describes constant acceleration that started with an initial nonzero velocity. To determine the displacement, you can divide the area under the graph into a rectangle and a triangle. The total area is then: If t i = 0: Motion with Constant Acceleration Copyright © McGraw-Hill Education Velocity with Average Acceleration Position with Average Acceleration

Rearrange the velocity with average equation to solve for t f = (v f − v i )/a. Substitute into the equation for position with average acceleration equation and rearrange. Motion with Constant Acceleration Copyright © McGraw-Hill Education An Alternate Equation Velocity with Constant Acceleration

Motion with Constant Acceleration Copyright © McGraw-Hill Education An Alternate Equation Response SKETCH AND ANALYZE THE PROBLEM Sketch the situation. List the knowns and unknowns. SOLVE FOR THE UNKNOWN Use the relationship among final velocity, initial velocity, acceleration, and displacement. EVALUATE THE ANSWER 2.1 m/s is a reasonable speed for jogging. Use with Example Problem 4. Problem Joni jogs at a velocity of 2.50 m/s. If she then accelerates at a constant − 0.10 m/s 2, how fast will she be jogging when she has moved 10.0 m? vivi a KNOWNUNKNOWN v i = 2.50 m/sv f = ? a = − 0.10 m/s 2 Δx = 10.0 m +x+x

Motion with Constant Acceleration Copyright © McGraw-Hill Education An Alternate Equation SOLVE FOR THE UNKNOWN For the first interval, use the relationship among velocity, time, and displacement. For the second interval, use the relationship among distance, initial and final velocity, and acceleration. Use with Example Problem 5. Problem A cat runs at 2.0 m/s for 3.0 s, then slows to a stop with an acceleration of −0.80 m/s 2. What is the cat’s displacement during this movement? Response SKETCH AND ANALYZE THE PROBLEM Sketch the situation. List the knowns and unknowns. KNOWNUNKNOWN v i = 2.0 m/sΔx 1 = ? Δt 1 = 3.0 sΔx 2 = ? a = − 0.80 m/s 2 Δx = ? v f = 0 Part 1 vivi vivi a Part 2 +x+x +x+x

Motion with Constant Acceleration Copyright © McGraw-Hill Education An Alternate Equation SOLVE FOR THE UNKNOWN Add the two displacements together. Use with Example Problem 5. Problem A cat runs at 2.0 m/s for 3.0 s, then slows to a stop with an acceleration of −0.80 m/s 2. What is the cat’s displacement during this movement? Response SKETCH AND ANALYZE THE PROBLEM Sketch the situation. List the knowns and unknowns.. KNOWNUNKNOWN v i = 2.0 m/sΔx 1 = 6.0 m Δt 1 = 3.0 sΔx 2 = 2.5 m a = − 0.80 m/s 2 Δx = ? v f = 0 EVALUATE THE ANSWER 8.5 meters is a reasonable distance for a cat to run. Part 1 vivi vivi a Part 2 +x+x +x+x

Free Fall Copyright © McGraw-Hill Education Galileo’s Discovery and Free-Fall Acceleration

Free fall is the motion of an object when gravity is the only significant force acting on it. The acceleration of an object due only to the effect of gravity is known as free-fall acceleration. Free-fall acceleration is represented by the symbol g. Near Earth’s surface, g = 9.80 m/s 2 downward. Free Fall Copyright © McGraw-Hill Education Free-Fall Acceleration