SLIDE 1
Constant Acceleration January 28, 2013 - p. 1/7
Constant Acceleration January 28, 2013 - p. 2/7
January 28, Week 3 Today: Chapter 2, Constant Accleration Homework - - PowerPoint PPT Presentation
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
SLIDE 2
SLIDE 3
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
SLIDE 4
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.
SLIDE 5
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.
SLIDE 6
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, which
- f the following are correct signs for its kinematical quantities?
b b b b
SLIDE 7
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, which
- f the following are correct signs for its kinematical quantities?
x
b b b b b b b b
SLIDE 8
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, which
- f the following are correct signs for its kinematical quantities?
x
b b b b
x vx ax
b b b b
SLIDE 9
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, which
- f the following are correct signs for its kinematical quantities?
x
b b b b
x vx ax (a) − + +
b b b b
SLIDE 10
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, which
- f the following are correct signs for its kinematical quantities?
x
b b b b
x vx ax (a) − + + (b) − + −
b b b b
SLIDE 11
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, which
- f the following are correct signs for its kinematical quantities?
x
b b b b
x vx ax (a) − + + (b) − + − (c) − − +
b b b b
SLIDE 12
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, which
- f the following are correct signs for its kinematical quantities?
x
b b b b
x vx ax (a) − + + (b) − + − (c) − − + (d) − − −
b b b b
SLIDE 13
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, which
- f the following are correct signs for its kinematical quantities?
x
b b b b
x vx ax (a) − + + (b) − + − (c) − − + (d) − − − (e) + − +
b b b b
SLIDE 14
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, which
- f the following are correct signs for its kinematical quantities?
x
b b b b
x vx ax (a) − + + (b) − + − (c) − − + (d) − − − (e) + − +
SLIDE 15
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration:
SLIDE 16
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration: t ax Acceleration versus time
SLIDE 17
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration: t ax Acceleration versus time
SLIDE 18
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration: t ax Acceleration versus time t v Velocity versus time
SLIDE 19
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration: t ax Acceleration versus time t v Velocity versus time
SLIDE 20
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration: t ax Acceleration versus time t v Velocity versus time t x Position versus time
SLIDE 21
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration: t ax Acceleration versus time t v Velocity versus time t x Position versus time
SLIDE 22
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration: t ax Acceleration versus time t v Velocity versus time v1 t1 v2 t2 t x Position versus time
SLIDE 23
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration: t ax Acceleration versus time t v Velocity versus time v1 t1 v2 t2 t x Position versus time (vx)2 = (vx)1 + ax∆t
SLIDE 24
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration: t ax Acceleration versus time t v Velocity versus time v1 t1 v2 t2 t x Position versus time x1 t1 x2 t2 (vx)2 = (vx)1 + ax∆t
SLIDE 25
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration: t ax Acceleration versus time t v Velocity versus time v1 t1 v2 t2 t x Position versus time x1 t1 x2 t2 (vx)2 = (vx)1 + ax∆t x2 = x1 + (vx)1 ∆t + 1 2ax (∆t)2
SLIDE 26
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration: t ax Acceleration versus time t v Velocity versus time v1 t1 v2 t2 t x Position versus time x1 t1 x2 t2 (vx)2 = (vx)1 + ax∆t x2 = x1 + (vx)1 ∆t + 1 2ax (∆t)2 (vx)2
2 = (vx)2 1 + 2ax∆x ← From Algebra
SLIDE 27
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.
SLIDE 28
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. t v Velocity versus time v1 v2 t2 t x Position versus time x1 x2 t2
SLIDE 29
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. t v Velocity versus time v1 v2 t2 t x Position versus time x1 x2 t2 t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t
SLIDE 30
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. t v Velocity versus time v1 v2 t t x Position versus time x1 x2 t t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t
SLIDE 31
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. t v Velocity versus time v1 v2 t t x Position versus time x1 x2 t t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t Better Notation: v2 = vx, x2 = x v1 = v0x, x1 = x0
SLIDE 32
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. t v Velocity versus time v0x vx t t x Position versus time x1 x2 t t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t Better Notation: v2 = vx, x2 = x v1 = v0x, x1 = x0
SLIDE 33
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. t v Velocity versus time v0x vx t t x Position versus time x0 x t t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t Better Notation: v2 = vx, x2 = x v1 = v0x, x1 = x0
SLIDE 34
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. t v Velocity versus time v0x vx t t x Position versus time x0 x t t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t Better Notation: v2 = vx, x2 = x v1 = v0x, x1 = x0 (vx)2 = (vx)1 + ax∆t x2 = x1 + (vx)1 ∆t + 1 2ax (∆t)2 (vx)2
2 = (vx)2 1 + 2ax∆x
SLIDE 35
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. t v Velocity versus time v0x vx t t x Position versus time x0 x t t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t Better Notation: v2 = vx, x2 = x v1 = v0x, x1 = x0 vx = v0x + axt x2 = x1 + (vx)1 ∆t + 1 2ax (∆t)2 (vx)2
2 = (vx)2 1 + 2ax∆x
SLIDE 36
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. t v Velocity versus time v0x vx t t x Position versus time x0 x t t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t Better Notation: v2 = vx, x2 = x v1 = v0x, x1 = x0 vx = v0x + axt x = x0 + (v0x)t + 1 2axt2 (vx)2
2 = (vx)2 1 + 2ax∆x
SLIDE 37
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. t v Velocity versus time v0x vx t t x Position versus time x0 x t t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t Better Notation: v2 = vx, x2 = x v1 = v0x, x1 = x0 vx = v0x + axt x = x0 + (v0x)t + 1 2axt2 v2
x = v2 0x + 2ax (x − x0)
SLIDE 38
Constant Acceleration January 28, 2013 - p. 6/7
Example I
x = x0 + (v0x)t + 1 2axt2 vx = v0x + axt v2
x = v2 0x + 2ax (x − x0)
SLIDE 39
Constant Acceleration January 28, 2013 - p. 6/7
Example I
x = x0 + (v0x)t + 1 2axt2 vx = v0x + axt v2
x = v2 0x + 2ax (x − x0)
Example: A car is traveling on a straight road with a speed of 30.0 m/s when the driver hits the brakes causing a constant deceleration of 2.5 m/s2. How long does it take and how far does the car go while stopping?
SLIDE 40
Constant Acceleration January 28, 2013 - p. 7/7
Problem Solving Exercise
A bicyclist is stopped at a red light. When the light turns green, he accelerates at 0.67 m/s2. To find how long it takes him to cross the 3.2-m-long intersection, we would use which equation of motion?
SLIDE 41
Constant Acceleration January 28, 2013 - p. 7/7
Problem Solving Exercise
A bicyclist is stopped at a red light. When the light turns green, he accelerates at 0.67 m/s2. To find how long it takes him to cross the 3.2-m-long intersection, we would use which equation of motion? (a) x = x0 + (v0x)t + 1 2axt2
SLIDE 42
Constant Acceleration January 28, 2013 - p. 7/7
Problem Solving Exercise
A bicyclist is stopped at a red light. When the light turns green, he accelerates at 0.67 m/s2. To find how long it takes him to cross the 3.2-m-long intersection, we would use which equation of motion? (a) x = x0 + (v0x)t + 1 2axt2 (b) vx = v0x + axt
SLIDE 43
Constant Acceleration January 28, 2013 - p. 7/7
Problem Solving Exercise
A bicyclist is stopped at a red light. When the light turns green, he accelerates at 0.67 m/s2. To find how long it takes him to cross the 3.2-m-long intersection, we would use which equation of motion? (a) x = x0 + (v0x)t + 1 2axt2 (b) vx = v0x + axt (c) v2
x = v2 0x + 2ax (x − x0)
SLIDE 44
Constant Acceleration January 28, 2013 - p. 7/7
Problem Solving Exercise
A bicyclist is stopped at a red light. When the light turns green, he accelerates at 0.67 m/s2. To find how long it takes him to cross the 3.2-m-long intersection, we would use which equation of motion? (a) x = x0 + (v0x)t + 1 2axt2 (b) vx = v0x + axt (c) v2
x = v2 0x + 2ax (x − x0)
(d) v = ∆x ∆t
SLIDE 45
Constant Acceleration January 28, 2013 - p. 7/7
Problem Solving Exercise
A bicyclist is stopped at a red light. When the light turns green, he accelerates at 0.67 m/s2. To find how long it takes him to cross the 3.2-m-long intersection, we would use which equation of motion? (a) x = x0 + (v0x)t + 1 2axt2 (b) vx = v0x + axt (c) v2
x = v2 0x + 2ax (x − x0)
(d) v = ∆x ∆t (e) None of these.
SLIDE 46
Constant Acceleration January 28, 2013 - p. 7/7
Problem Solving Exercise
A bicyclist is stopped at a red light. When the light turns green, he accelerates at 0.67 m/s2. To find how long it takes him to cross the 3.2-m-long intersection, we would use which equation of motion? (a) x = x0 + (v0x)t + 1 2axt2 (b) vx = v0x + axt (c) v2
x = v2 0x + 2ax (x − x0)