January 28, Week 3 Today: Chapter 2, Constant Accleration Homework - PowerPoint PPT Presentation

January 28, Week 3 Today: Chapter 2, Constant Accleration Homework Assignment #3 - Due February 1 Mastering Physics: 6 problems from chapter 2. Written Question: 2.88 Box numbers can be found on webpage Wednesday office hours will be 2:30-5:00 For now on, Mastering Physics will take off points for missed homework questions. Constant Acceleration January 28, 2013 - p. 1/7

Review Three physical quantities of kinematics: Constant Acceleration January 28, 2013 - p. 2/7

Review Three physical quantities of kinematics: Position, x - Where an object is located = how far and what direction from origin Constant Acceleration January 28, 2013 - p. 2/7

Review Three physical quantities of kinematics: Position, x - Where an object is located = how far and what direction from origin Velocity, v - How fast an object is going and direction motion = speed and from its current position what direction is it going towards. Also, slope of the position-versus-time graph. Constant Acceleration January 28, 2013 - p. 2/7

Review Three physical quantities of kinematics: Position, x - Where an object is located = how far and what direction from origin Velocity, v - How fast an object is going and direction motion = speed and from its current position what direction is it going towards. Also, slope of the position-versus-time graph. Acceleration, a - The rate at which velocity is changing. Has same sign as velocity for speeding up. Opposite sign for slowing down. Slope of the velocity-versus-time graph. Constant Acceleration January 28, 2013 - p. 2/7

b b b b Acceleration Exercise For the following motion diagram and coordinate system, which of the following are correct signs for its kinematical quantities? Constant Acceleration January 28, 2013 - p. 3/7

b b b b b b b b Acceleration Exercise For the following motion diagram and coordinate system, which of the following are correct signs for its kinematical quantities? x 0 Constant Acceleration January 28, 2013 - p. 3/7

b b b b b b b b Acceleration Exercise For the following motion diagram and coordinate system, which of the following are correct signs for its kinematical quantities? x 0 x v x a x Constant Acceleration January 28, 2013 - p. 3/7

b b b b b b b b Acceleration Exercise For the following motion diagram and coordinate system, which of the following are correct signs for its kinematical quantities? x 0 x v x a x (a) + + − Constant Acceleration January 28, 2013 - p. 3/7

b b b b b b b b Acceleration Exercise For the following motion diagram and coordinate system, which of the following are correct signs for its kinematical quantities? x 0 x v x a x (a) + + − (b) + − − Constant Acceleration January 28, 2013 - p. 3/7

b b b b b b b b Acceleration Exercise For the following motion diagram and coordinate system, which of the following are correct signs for its kinematical quantities? x 0 x v x a x (a) + + − (b) + − − (c) + − − Constant Acceleration January 28, 2013 - p. 3/7

b b b b b b b b Acceleration Exercise For the following motion diagram and coordinate system, which of the following are correct signs for its kinematical quantities? x 0 x v x a x (a) + + − (b) + − − (c) + − − (d) − − − Constant Acceleration January 28, 2013 - p. 3/7

b b b b b b b b Acceleration Exercise For the following motion diagram and coordinate system, which of the following are correct signs for its kinematical quantities? x 0 x v x a x (a) + + − (b) + − − (c) + − − (d) − − − (e) + + − Constant Acceleration January 28, 2013 - p. 3/7

b b b b Acceleration Exercise For the following motion diagram and coordinate system, which of the following are correct signs for its kinematical quantities? x 0 x v x a x (a) + + − (b) + − − (c) + − − (d) − − − (e) + + − Constant Acceleration January 28, 2013 - p. 3/7

Constant Acceleration For a constant acceleration: Constant Acceleration January 28, 2013 - p. 4/7

Constant Acceleration For a constant acceleration: a x Acceleration versus time t Constant Acceleration January 28, 2013 - p. 4/7

Constant Acceleration For a constant acceleration: a x Acceleration versus time t Constant Acceleration January 28, 2013 - p. 4/7

Constant Acceleration For a constant acceleration: a x v Velocity versus time Acceleration versus time t t Constant Acceleration January 28, 2013 - p. 4/7

Constant Acceleration For a constant acceleration: a x v Velocity versus time Acceleration versus time t t Constant Acceleration January 28, 2013 - p. 4/7

Constant Acceleration For a constant acceleration: a x v Velocity versus time Acceleration versus time t t x Position versus time t Constant Acceleration January 28, 2013 - p. 4/7

Constant Acceleration For a constant acceleration: a x v Velocity versus time Acceleration versus time t t x Position versus time t Constant Acceleration January 28, 2013 - p. 4/7

Constant Acceleration For a constant acceleration: a x v Velocity versus time Acceleration versus time v 2 v 1 t t t 1 t 2 x Position versus time t Constant Acceleration January 28, 2013 - p. 4/7

Constant Acceleration For a constant acceleration: a x v Velocity versus time Acceleration versus time v 2 v 1 t t t 1 t 2 x Position versus time ( v x ) 2 = ( v x ) 1 + a x ∆ t t Constant Acceleration January 28, 2013 - p. 4/7

Constant Acceleration For a constant acceleration: a x v Velocity versus time Acceleration versus time v 2 v 1 t t t 1 t 2 x Position versus time ( v x ) 2 = ( v x ) 1 + a x ∆ t x 2 x 1 t t 1 t 2 Constant Acceleration January 28, 2013 - p. 4/7

Constant Acceleration For a constant acceleration: a x v Velocity versus time Acceleration versus time v 2 v 1 t t t 1 t 2 x Position versus time ( v x ) 2 = ( v x ) 1 + a x ∆ t x 2 x 2 = x 1 + ( v x ) 1 ∆ t + 1 2 a x (∆ t ) 2 x 1 t t 1 t 2 Constant Acceleration January 28, 2013 - p. 4/7

Constant Acceleration For a constant acceleration: a x v Velocity versus time Acceleration versus time v 2 v 1 t t t 1 t 2 x Position versus time ( v x ) 2 = ( v x ) 1 + a x ∆ t x 2 x 2 = x 1 + ( v x ) 1 ∆ t + 1 2 a x (∆ t ) 2 x 1 ( v x ) 2 2 = ( v x ) 2 t 1 + 2 a x ∆ x ← From Algebra t 1 t 2 Constant Acceleration January 28, 2013 - p. 4/7

A Simplification We can make our equations looks a little simpler by always assuming that the initial time is zero. Constant Acceleration January 28, 2013 - p. 5/7

A Simplification We can make our equations looks a little simpler by always assuming that the initial time is zero. v x Velocity versus time Position versus time x 2 v 2 v 1 x 1 t t t 2 t 2 Constant Acceleration January 28, 2013 - p. 5/7

A Simplification We can make our equations looks a little simpler by always assuming that the initial time is zero. v x Velocity versus time Position versus time x 2 v 2 v 1 x 1 t t t 2 t 2 t 1 = 0 ⇒ ∆ t = t 2 − 0 = t 2 = t Constant Acceleration January 28, 2013 - p. 5/7

A Simplification We can make our equations looks a little simpler by always assuming that the initial time is zero. v x Velocity versus time Position versus time x 2 v 2 v 1 x 1 t t t t t 1 = 0 ⇒ ∆ t = t 2 − 0 = t 2 = t Constant Acceleration January 28, 2013 - p. 5/7

A Simplification We can make our equations looks a little simpler by always assuming that the initial time is zero. v x Velocity versus time Position versus time x 2 v 2 v 1 x 1 t t t t v 2 = v x , x 2 = x t 1 = 0 ⇒ ∆ t = t 2 − 0 = t 2 = t Better Notation: v 1 = v 0 x , x 1 = x 0 Constant Acceleration January 28, 2013 - p. 5/7

A Simplification We can make our equations looks a little simpler by always assuming that the initial time is zero. v x Velocity versus time Position versus time x 2 v x v 0 x x 1 t t t t v 2 = v x , x 2 = x t 1 = 0 ⇒ ∆ t = t 2 − 0 = t 2 = t Better Notation: v 1 = v 0 x , x 1 = x 0 Constant Acceleration January 28, 2013 - p. 5/7

A Simplification We can make our equations looks a little simpler by always assuming that the initial time is zero. v x Velocity versus time Position versus time x v x v 0 x x 0 t t t t v 2 = v x , x 2 = x t 1 = 0 ⇒ ∆ t = t 2 − 0 = t 2 = t Better Notation: v 1 = v 0 x , x 1 = x 0 Constant Acceleration January 28, 2013 - p. 5/7

A Simplification We can make our equations looks a little simpler by always assuming that the initial time is zero. v x Velocity versus time Position versus time x v x v 0 x x 0 t t t t v 2 = v x , x 2 = x t 1 = 0 ⇒ ∆ t = t 2 − 0 = t 2 = t Better Notation: v 1 = v 0 x , x 1 = x 0 x 2 = x 1 + ( v x ) 1 ∆ t + 1 2 a x (∆ t ) 2 ( v x ) 2 = ( v x ) 1 + a x ∆ t ( v x ) 2 2 = ( v x ) 2 1 + 2 a x ∆ x Constant Acceleration January 28, 2013 - p. 5/7

A Simplification We can make our equations looks a little simpler by always assuming that the initial time is zero. v x Velocity versus time Position versus time x v x v 0 x x 0 t t t t v 2 = v x , x 2 = x t 1 = 0 ⇒ ∆ t = t 2 − 0 = t 2 = t Better Notation: v 1 = v 0 x , x 1 = x 0 x 2 = x 1 + ( v x ) 1 ∆ t + 1 2 a x (∆ t ) 2 v x = v 0 x + a x t ( v x ) 2 2 = ( v x ) 2 1 + 2 a x ∆ x Constant Acceleration January 28, 2013 - p. 5/7