Upload
stephanie-goodwin
View
219
Download
1
Embed Size (px)
Citation preview
Motion in One Dimension
Unit 1
Lesson 1 : Position, Velocity, and Speed
Position : location of a particle with respect to a chosen reference point
Displacement : the change in position in some time interval
x = xf – xi
Distance : the length of a path followed by a particle
Displacement is a VECTOR
QUANTITY
Distance is a SCALAR
QUANTITY
Average Velocity : a particle’s displacement (x) divided by the time interval (t)
during which that displacement occurs
V =x
tAverage Speed : the total distance traveled divided
by the total time interval required to travel that distance
Avg speed =total distance
total time
Avg Velocity is a VECTOR
QUANTITY
Avg Speed is a SCALAR
QUANTITY
Example 1
Position t(s) x(m)
A 0 30
B 10 52
C 20 38
D 30 0
E 40 -37
F 50 -53
Find the displacement, average velocity, and average speed of the object between
positions A and F.
Lesson 2 : Instantaneous Velocity and Speed
Instantaneous Velocity : the limiting value of the ratio x/t as t approaches zero
v =x
tlim
t 0
Instantaneous Speed : the magnitude of the instantaneous velocity
Example 1
A) A ball thrown directly upward rises to a highest point and falls back in the thrower’s hand.
B) A race car starts from rest and speeds up to 100 m/s.
C) A spacecraft drifts through space at constant velocity.
Are there any points at which the instantaneous velocity has the same value as the average velocity over the entire motion ? If so, identify the point(s).
What is the instantaneous velocity at t = 5.0 s ?
Answer : Slope of the tangent line drawn at the time in question.
Example 2
Lesson 3 : AccelerationAverage Acceleration : the change in velocity
(v) divided by the time interval (t) during which that change occurs
a =v
t=
vf – vi
tf – ti
Instantaneous Acceleration : the limit of the average acceleration as t appoaches zero
a =v
tlim
t 0
Example 1
What is the instantaneous acceleration at t = 2.0 s ?
Answer : Slope of the tangent line drawn at the time in question.
Example 2
The velocity of a particle moving along the x-axis varies in time according to the expression
vx = (40 – 5t2) m/s, where t is in seconds.
A) Find the average acceleration in the time interval t =0 to t = 2.0 s.
B) Determine the acceleration at t = 2.0 s.
Lesson 4 : Motion Diagrams
v
No acc
Motion A
v
a
Motion B
v
a
Motion C
Graphs for each motion :
x
Motion A
t
v
t
at
x
Motion B
t
v
t
at
x
Motion C
t
v
t
at
Example 1
Draw the corresponding x vs. t and a vs. t graphs.
A)v
t
x
t
a
t
B)v
t
x
t
a
t
vC)
t
x
t
a
t
Example 2
Draw the corresponding v vs. t and a vs. t graphs.
A)x
t
v
t
a
t
B)x
t
v
t
a
t
xC)
t
v
t
a
t
What is the displacement of the object from t = 0 to 3 s ?
Answer : The area under the graph equals displacement
Example 3
Negative Area :
Object is moving toward smaller x
values and displacement is
decreasing
Lesson 5 : Kinematic Equations
a =v
t=
vf – vi
tf – ti
vf = vi + at
vf = vi + at (for constant a)
v =vi + vf
2(for constant a)
x = xf - xi
x = vt
xf – xi = vt
xf – xi = ½ (vi + vf)t
xf = xi + ½ (vi + vf)t (for constant a)
xf = xi + ½ (vi + vf)t
xf = xi + ½ [vi + (vi +at)]t
vf = vi + at
xf = xi + vit + ½ at2 (for constant a)
xf = xi + ½ (vi + vf)t
xf = xi + ½ (vi + vf) (vf - vi
a)
vf2 = vi
2 + 2a (xf – xi) (for constant a)
Summary
vf = vi + at (for constant a)
v =vi + vf
2(for constant a)
xf = xi + ½ (vi + vf)t (for constant a)
xf = xi + vit + ½ at2 (for constant a)
vf2 = vi
2 + 2a (xf – xi) (for constant a)
Example 1
A jet lands on an aircraft carrier at 140 mph (63 m/s).
A) What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the airplane and brings it to a stop ?
B) If the plane touches down at position xi = 0, what is the final position of the plane ?
Example 2
A car traveling at a constant speed of 45.0 m/s passes a trooper hidden behind
a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to
catch it, accelerating at a constant rate of 3.00 m/s2. How long does it take her to
overtake the car ?
Lesson 6 : Freely Falling Objects
Neglecting air resistance, all objects dropped near the Earth’s surface fall
toward the Earth with the same constant acceleration under the influence of the
Earth’s gravity.
A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial
motion.
Free-fall acceleration (g) = 9.80 m/s2
+y -y
ay = -g = -9.80 m/s2
For making quick estimates, use g = 10 m/s2
Example 1
A ball is tossed straight up at 25 m/s. Estimate its velocity at 1 s intervals.
Time (s) Velocity (m/s)
0 +25
1
2
3
4
5
Example 2
A stone is thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward.
The building is 50.0 m high, and the stone just misses the edge of the roof on its way down.
A) Determine the time at which the stone reaches its maximum height.
B) Determine the maximum height.
D) Determine the velocity of the stone at this instant.
E) Determine the velocity and position of the stone at t = 5.00 s.
C) Determine the time at which the stone returns to the height from which it was thrown.
Lesson 7 : Using Calculus in Kinematics
Instantaneous Velocity v =x
tlim
t 0
This limit is called the derivative of x with respect to t.
v =x
tlim
t 0
=dx
dt
v =dx
dt
f(x) = axn
Derivative of Power Function
dfdx
= naxn-1
Example 1
An object is moving in one dimension according to the formula x(t) = 2t3 + t2 – 4.
Find its velocity at t = 2 s.
Instantaneous Acceleration a =v
tlim
t 0
This limit is called the derivative of v with respect to t.
a =v
tlim
t 0
=dv
dt
a =dv
dt
Example 2
An object is moving in one dimension according to the formula v(t) = (40 – 5t2).
Find its acceleration at t = 2 s.
Acceleration is also the second derivative of position.
a =v
tlim
t 0
=dv
dt=
d2x
dt2
Example 3
An object is moving in one dimension according to the formula x(t) = 12 – 4t + 2t3.
Find its acceleration at t = 3 s.
The Integral or Antiderivative
v
tti tf
tn
vn
xn = vntn = shaded area
x = vntn = total area
As rectangles get narrower, tn 0
x = limtn 0
vn tn
Displacement = area under v-t graph
This is called the Definite Integral
limtn 0
vn tn = v(t) dtti
tf
Integrating Velocity
v(t) =dx
dt
dx = v dt
xi
xf
dx =ti
tf
v dt
xf – xi = ti
tf
v dt
Integrating Acceleration
a(t) =dv
dt
dv = a dt
vi
vf
dv =ti
tf
a dt
vf – vi = ti
tf
a dt
Finding Antiderivatives
If k is a constant,
k dx = kx + C
xn dx =xn+1
n + 1+ C (n = 1)
Example 4
A car currently moving at 10. m/s accelerates non-uniformly according to
a(t) = 3t2. Find its velocity at t = 2 s.
Example 5
An object is moving according to the formula v(t) = 5t2 – 4t. If the object starts
from rest, find its position at t = 5s .
Example 6
An object is moving according to the formula a(t) = 2t – 4, with an initial velocity of + 4 m/s, and an initial position x = 0 at
time t = 0. Find the position and velocity at arbitrary times.