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Constant Acceleration January 28, 2013 - p. 1/7
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 missedhomework questions.
Constant Acceleration January 28, 2013 - p. 2/7
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 whatdirection 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 whatdirection from origin
Velocity, v - How fast an object is going and direction motion =speed and from its current position what direction is it goingtowards. 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 whatdirection from origin
Velocity, v - How fast an object is going and direction motion =speed and from its current position what direction is it goingtowards. Also, slope of the position-versus-time graph.
Acceleration, a - The rate at which velocity is changing. Hassame sign as velocity for speeding up. Opposite sign forslowing down. Slope of the velocity-versus-time graph.
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, whichof the following are correct signs for its kinematical quantities?
b b b b
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, whichof the following are correct signs for its kinematical quantities?
x0
b b b b
b b b b
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, whichof the following are correct signs for its kinematical quantities?
x0
b b b b
x vx ax
b b b b
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, whichof the following are correct signs for its kinematical quantities?
x0
b b b b
x vx ax
(a) − + +
b b b b
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, whichof the following are correct signs for its kinematical quantities?
x0
b b b b
x vx ax
(a) − + +
(b) − + −
b b b b
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, whichof the following are correct signs for its kinematical quantities?
x0
b b b b
x vx ax
(a) − + +
(b) − + −
(c) − − +
b b b b
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, whichof the following are correct signs for its kinematical quantities?
x0
b b b b
x vx ax
(a) − + +
(b) − + −
(c) − − +
(d) − − −
b b b b
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, whichof the following are correct signs for its kinematical quantities?
x0
b b b b
x vx ax
(a) − + +
(b) − + −
(c) − − +
(d) − − −
(e) + − +
b b b b
Constant Acceleration January 28, 2013 - p. 3/7
Acceleration Exercise
For the following motion diagram and coordinate system, whichof the following are correct signs for its kinematical quantities?
x0
b b b b
x vx ax
(a) − + +
(b) − + −
(c) − − +
(d) − − −
(e) + − +
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration:
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration:
t
ax Acceleration versus time
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration:
t
ax Acceleration versus time
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration:
t
ax Acceleration versus time
t
v Velocity versus time
Constant Acceleration January 28, 2013 - p. 4/7
Constant Acceleration
For a constant acceleration:
t
ax Acceleration versus time
t
v Velocity versus time
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
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
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
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
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
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
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
Constant Acceleration January 28, 2013 - p. 5/7
A Simplification
We can make our equations looks a little simpler by alwaysassuming 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 alwaysassuming that the initial time is zero.
t
v Velocity versus time
v1
v2
t2t
x Position versus time
x1
x2
t2
Constant Acceleration January 28, 2013 - p. 5/7
A Simplification
We can make our equations looks a little simpler by alwaysassuming that the initial time is zero.
t
v Velocity versus time
v1
v2
t2t
x Position versus time
x1
x2
t2
t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t
Constant Acceleration January 28, 2013 - p. 5/7
A Simplification
We can make our equations looks a little simpler by alwaysassuming that the initial time is zero.
t
v Velocity versus time
v1
v2
tt
x Position versus time
x1
x2
t
t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t
Constant Acceleration January 28, 2013 - p. 5/7
A Simplification
We can make our equations looks a little simpler by alwaysassuming that the initial time is zero.
t
v Velocity versus time
v1
v2
tt
x Position versus time
x1
x2
t
t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t Better Notation:v2 = vx, x2 = x
v1 = v0x, x1 = x0
Constant Acceleration January 28, 2013 - p. 5/7
A Simplification
We can make our equations looks a little simpler by alwaysassuming that the initial time is zero.
t
v Velocity versus time
v0x
vx
tt
x Position versus time
x1
x2
t
t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t Better Notation:v2 = vx, x2 = x
v1 = v0x, x1 = x0
Constant Acceleration January 28, 2013 - p. 5/7
A Simplification
We can make our equations looks a little simpler by alwaysassuming that the initial time is zero.
t
v Velocity versus time
v0x
vx
tt
x Position versus time
x0
x
t
t1 = 0 ⇒ ∆t = t2 − 0 = t2 = t Better Notation:v2 = vx, x2 = x
v1 = v0x, x1 = x0
Constant Acceleration January 28, 2013 - p. 5/7
A Simplification
We can make our equations looks a little simpler by alwaysassuming that the initial time is zero.
t
v Velocity versus time
v0x
vx
tt
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
Constant Acceleration January 28, 2013 - p. 5/7
A Simplification
We can make our equations looks a little simpler by alwaysassuming that the initial time is zero.
t
v Velocity versus time
v0x
vx
tt
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
Constant Acceleration January 28, 2013 - p. 5/7
A Simplification
We can make our equations looks a little simpler by alwaysassuming that the initial time is zero.
t
v Velocity versus time
v0x
vx
tt
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
2axt
2
(vx)2
2= (vx)
2
1+ 2ax∆x
Constant Acceleration January 28, 2013 - p. 5/7
A Simplification
We can make our equations looks a little simpler by alwaysassuming that the initial time is zero.
t
v Velocity versus time
v0x
vx
tt
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
2axt
2
v2x= v2
0x+ 2ax (x− x0)
Constant Acceleration January 28, 2013 - p. 6/7
Example I
x = x0 + (v0x)t+1
2axt
2 vx = v0x + axt
v2x= v2
0x+ 2ax (x− x0)
Constant Acceleration January 28, 2013 - p. 6/7
Example I
x = x0 + (v0x)t+1
2axt
2 vx = v0x + axt
v2x= v2
0x+ 2ax (x− x0)
Example: A car is traveling on a straight road with a speed of30.0m/s when the driver hits the brakes causing a constantdeceleration of 2.5m/s2. How long does it take and how fardoes the car go while stopping?
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 0.67m/s2. To find how long it takes him tocross the 3.2-m-long intersection, we would use whichequation of motion?
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 0.67m/s2. To find how long it takes him tocross the 3.2-m-long intersection, we would use whichequation of motion?
(a) x = x0 + (v0x)t+1
2axt
2
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 0.67m/s2. To find how long it takes him tocross the 3.2-m-long intersection, we would use whichequation of motion?
(a) x = x0 + (v0x)t+1
2axt
2
(b) vx = v0x + axt
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 0.67m/s2. To find how long it takes him tocross the 3.2-m-long intersection, we would use whichequation of motion?
(a) x = x0 + (v0x)t+1
2axt
2
(b) vx = v0x + axt
(c) v2x= v2
0x+ 2ax (x− x0)
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 0.67m/s2. To find how long it takes him tocross the 3.2-m-long intersection, we would use whichequation of motion?
(a) x = x0 + (v0x)t+1
2axt
2
(b) vx = v0x + axt
(c) v2x= v2
0x+ 2ax (x− x0)
(d) v =∆x
∆t
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 0.67m/s2. To find how long it takes him tocross the 3.2-m-long intersection, we would use whichequation of motion?
(a) x = x0 + (v0x)t+1
2axt
2
(b) vx = v0x + axt
(c) v2x= v2
0x+ 2ax (x− x0)
(d) v =∆x
∆t
(e) None of these.
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 0.67m/s2. To find how long it takes him tocross the 3.2-m-long intersection, we would use whichequation of motion?
(a) x = x0 + (v0x)t+1
2axt
2
(b) vx = v0x + axt
(c) v2x= v2
0x+ 2ax (x− x0)
(d) v =∆x
∆t
(e) None of these.
x0 = 0, x = 3.2m
v0x = 0, ax = 0.67m/s2
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