Loading in 2 Seconds...
Loading in 2 Seconds...
Homework – due Friday , 9/23 Reading assignment: 2.1-2.6 Questions: 2, 7, 9, 12, 13, 22, 23, 24, 25, 29, 31, 34, 40, 44 – the solutions are on the school website. Homework – due Tuesday, 9/20 – 11:00 pm Mastering physics wk 3. Chapter 2. Motion along a straight line
Reading assignment: 2.1-2.6
Questions: 2, 7, 9, 12, 13, 22, 23, 24, 25, 29, 31, 34, 40, 44 – the solutions are on the school website.
Homework – due Tuesday, 9/20 – 11:00 pm
Mastering physics wk 3
Motion along a straight line
Using mathematics to describe motion in terms of position, velocity and acceleration.
Use concepts of ideal particle, only consider translational motion
Understand the relationship between Displacement, Time, and Average Velocity
∆tDistance, Time, and Average Speed
x2 – x1
vav-x = =
t2 – t1Displacement, Time, and Average Velocity
∆x = x2 – x1
Let's suppose Taylor covered each leg of her journey in one second. This means that the total time for her trip was 6 seconds.
Average speed = 12 m/ 6s = 2 m/s
Average velocity = 3 m / 6 s = 0.5 m/s
Direction of velocity is east
Position at t2 = 4.0 s
Position at t1 = 1.0 s
x2 = 277 m
x1 = 19 m
4, 1, 3, 5, 2
286 s, 1770 m, 1570 m
572 s, 3540 m, 3140 m
a) Draw a graph showing the positions of both Tim and Rick versus time.
b) Write two sentences explaining who wins and why.
c) How long does it take Rick to cover the distance D?
d) Find Rick's average speed for covering the distance D.
e) How long does it take Tim to cover the distance?
vx = lim =
When ∆x is positive, vx is positive
When ∆x is negative, vx is negative
Slope of the line P1P2 represents the average velocity v between t1 and t2.velocity on P-T graph
To get instantaneous velocity at P1, we pick a point Pi which is extremely close to P1:
Instantaneous velocity (velocity) as the limit as we let ∆t →0. it is equal to the slope of the tangent to the curve at the point.
The slope of the tangent at any point equals the velocity at that point.
R, P, Q = S
Given x = 2.1t + 2.80, graph x vs. t and v vs. t
Constant (uniform) motion
Average velocity = instantaneous velocity
aav-x = =
t2 – t1
∆t2.3 average and instantaneous acceleration
Velocity describes how fast a body’s position change with time.
Acceleration describes how fast a body’s velocity change, it tells how speed and direction of motion are changing.
x(t) = at3 – bt2 + ct - d,
where a = 3.6 m/s3, b = 5.0 m/s2; c = 6 m/s; and d = 7.0 m
(a) Find the instantaneous acceleration at t = 2.4 s.
(b) Find the average acceleration over the first 2.4 seconds.
a is in the same direction as v
a is in the opposite direction as v
On a x-t graph, the acceleration is given by the curvature of the graph.
Curves up from the point: acceleration is positive
straight or not curves up or down: acceleration is zero
Curves down: acceleration is negative
P: v is not change;
Q: v is zero, changing from pos. to neg., first decrease in pos. then increase in neg.,
R: v is neg., constant;
S: v is zero, changing from neg. to pos., first decrease in neg. then increase in pos.,
vx – vx0
x – x0
tDerive equations for motion with constant acceleration
(assume t0 = 0)
A horizontal line indicate the slope = 0, a = 0
Since ax = ∆v / ∆t; ∆v = ax ∙ ∆t which is represented by the area.
The area indicate the change in velocity during ∆t
Slope: indicate acceleration
Area indicate the change in displacement from time = 0 to t
X = 55 m
a. t = 10 s.
b. v = 30 m/s.
c. d = 150 m
Four possible vx-t graphs are shown for the two vehicles in example 2.5. which graph is correct?
If we ignore air friction and the effects due to the earth’s rotation, all objects fall and rise at the constant acceleration.
The constant acceleration of a freely falling body is called the acceleration due to gravity, and we use letter g to represent its magnitude. Near the earth’s surface g = 9.8 m/s/s = 32 ft/s/s
On the surface of the moon, g = 1.6 m/s/s
On the surface of the sun, g = 270 m/s/s
a = -g
x = -4.9 m; v = -9.8 m/s
t = 2.0 s
x = -19.6 m; v = -19.6 m/s
t = 3.0 s
x = -44.1 m; v = -29.4 m/s
(9.80 m/s/s)t2 – (15 m/s)(t) – (5.00 m) = 0
The graph above shows velocity v versus time t for an object in linear motion. Sketch a graph of position x versus time t for this object?