 Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Displacement: As an object is moved from one place to another it is “displaced.” Displacement.

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Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Displacement: As an object is moved from one place to another it is “displaced.” Displacement in one dimension can be represented as the change in position x. This is known as Δ x. Δ x = x f - x i 1 2 3 4 5 6 7 8 9 Example An object is originally located 2 feet from the origin and is moved to a position 8 meters from the origin. Δ x = x f - x i = 8m - 2m Δ x = 6m

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Displacement: Δ x = x f - x i In the previous the object is a “Point Mass.” We are ignoring the size of the object and treating it as if it only existed at its center of mass. In the real world objects have size. Consider a car 4m long parked between two positions 25m apart. If the car moves from the first position to the final, how far does it travel?

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Velocity: The displacement of an object over a set interval of time is called the average velocity. 1 2 3 4 5 6 7 8 9 Example If the object in the first example moves takes 3 seconds to make the trip, what is its average velocity? = 8m - 2m = 2m/s 3s

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Notation: Look at the equation of velocity that we are using. Note the “Δ” symbol. This is called delta. It is used to denote a change in that listed quantity. Δ x denote a change in the x axis. Also note the arrow over the v This indicates that velocity is a “vector.” A vector has magnitude and direction. A quantity with only magnitude is a scalar

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Velocity: What is the average velocity of an object that goes 4.0m in 3.0 seconds, then 5.0 more meters in 2.0 seconds and then 2.0 more meters in 2.0 seconds? Problem solving 1. Draw a picture of the problem 2. List what you know 3. Solve 0 1 2 3 4 5 6 7 8 9 10 11 Δ x = x f - x i = 11.0m - 0m = 11.0m Δ t = 3.0s +2.0s +2.0s = 7.0s = 11.0m = 1.6m.s 7.0s

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Velocity: What is the average velocity of an object that goes 4.0m in 3.0 seconds, then 5.0 more meters in 2.0 seconds and then 2.0 more meters in 2.0 seconds? Why is the answer 1.6m/.s and not 1.571 m/.s ? 0 1 2 3 4 5 6 7 8 9 10 11 Δ x = x f - x i = 11.0m - 0.0m = 11.0m Δ t = 3.0s +2.0s +2.0s = 7.0s = 11.0m = 1.6m/.s 7.0s Because the lowest significant figures of the terms is determines the significant figures of the answer.

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Velocity: Velocity may be represented as a graph for example x vs t 1 2 3 4 5 6 7 8 9 seconds Meters 11 10 9 8 7 6 5 4 3 2 1 Δ x = x f - x i = 11.0m - 0m = 11.0m Δ t = 3.0s +2.0s +2.0s = 7.0s = 11.0m = 1.6m.s 7.0s

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Instantaneous Velocity: Is the representation of velocity at an exact instant. Lim Δt 0

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Constant Acceleration: Acceleration is the change in velocity with respect to time. We are discussing “constant acceleration” which means that the rate of change remains the same through the entire time interval being described. Example In the first example, if the object was at rest to begin with and was subject to a constant acceleration what was its velocity after3 seconds? What was the acceleration? = 8m - 2m = 2m/s = v f + v i v f = 4m/s 3s 2

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Constant Acceleration: Acceleration is the change in velocity with respect to time. We are discussing “constant acceleration” which means that the rate of change remains the same through the entire time interval being described. Example In the first example, if the object was at rest to begin with and was subject to a constant acceleration what was its velocity after3 seconds? What was the acceleration? = 2.0m/s = 0.67m/s 2 3.0s

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Constant Acceleration: Note the unites: Displacement is in m, Velocity is m/s and Acceleration is m/s 2 Δ x = x f - x i

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Motion We can rearrange and combine the equations for displacement, velocity and acceleration. Note – we discard the Δ t notation and just use t to denote the elapsed time. vf-vivf-vi Δ x = Δ v t = ( v f - v i )t Δt Δt v f = v i + a Δ t v f = v i + at 2 1/2 ( v f - v i )t Substitute v f = v i + at into the equation Δ x = 1/2 ( v i + v i + at)t x f = x i + v i t + at 2 2 1

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Motion Lastely -- v 2 f = v 2 i + 2 a Δ x

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Motion Know these three equations: v 2 f = v 2 i + 2 a Δ x x f = x i + v i t + at 2 2 1 v f = v i + at

Phys 220 Chapter 2: Motion in one Dimension 1/14/2015Phys 220 Motion https://login.cengage.com/cb/ E-TWQNSTPQ2CFDE The power point at the side is the homework for Week one. If you do not already have cengage set up you may down load this file and turn the HW in manually. Below are web address and course number for this class. https://login.cengage.com/cb/ E-TWQNSTPQ2CFDE

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