A skydiver is falling down (y component
negative) at a speed of 48 m/s. She
opens her chute. 11 seconds later, she
has slowed down to 26 m/s. Assume
her acceleration is constant. Draw
graphs for:
a. Acceleration (numerically correct)
b. Velocity (numerically correct)
c. Position (just the shape – assume the
ground is at zero)
What will the fan cart do when started with
these parameters (a = +2m/s2, v0 = -5m/s)?
A. Speed up
to the left
B. Speed up
to the right
C. Slow down
to the left
D. Slow down
to the right
Draw the graphs for your prediction (x0 = 0)
One dimensional constant acceleration
http://bcs.wiley.com/hebcs/Books?action=mininav&bcsId=4678&itemId=04
70223553&assetId=160341&resourceId=15300
Motion with Constant Acceleration
A jet plane has an acceleration of 2m/s/s, starting from
rest, at the origin. Fill in the table. Don’t use
kinematic equations:
t (s)
0
1
2
3
4
velocity (m/s)
Motion with Constant Acceleration
A jet plane has an acceleration of 2m/s/s, starting from rest, at the
origin. After 1 s, what is the position of the object?
t (s)
velocity, v (m/s)
position, x (m)
0
0
0
1
2
X
A. 2m
B. 1 m
C. Somewhere between 1m and 2m
D. Can’t be determined without more information
Motion with Constant Acceleration
 (v0  v)  t
x 
2
Δt is the time elapsed between v0 and v1.
Motion with Constant Acceleration
A jet plane has an acceleration of 2m/s/s, starting from rest, at the
origin. Fill in the table.
t (s)
velocity, v (m/s)
0
0
1
2
2
4
3
6
4
8
position, x (m)
Constant Velocity vs. Constant Acceleration
http://bcs.wiley.com/hebcs/Books?action=mininav&bcsId=4678&itemId=04
70223553&assetId=160341&resourceId=15300
After 1 second, which ball has gone further, the
one with constant velocity (red) or the one with
constant acceleration?
Constant Velocity vs. Constant Acceleration
http://bcs.wiley.com/hebcs/Books?action=mininav&bcsId=4678&itemId=04
70223553&assetId=160341&resourceId=15300
Go to www. Wiley.com, physics and astronomy,
introductory physics(algebra-based) keyword cutnell
Go to student companion site
Mute the audio please!!
Prove the following, using math: At what time do the
balls meet? At what position?
Pumping Blood – Example #1
The left ventricle of the
heart, each time it
pumps, accelerates
blood from rest to a
velocity of 26 cm/s. It
takes 0.15 s for the heart
to go through one cycle.
How far (2 sig figs)
does the blood travel?
a. 3.9 cm
b. 2.0 cm
Pumping Blood - Example #1
The left ventricle of the heart, each time it pumps,
accelerates blood from rest to a velocity of 26 cm/s.
It takes 0.15 s for the heart to go through one cycle.
How far (2 sig figs) does the blood travel? 2.0 cm
What is the acceleration that the heart provides?
Example Problem #2 – turning point
Bob throws the ball straight up at 20 m/s releasing the
ball 1.5 m above the ground.
a. What is the maximum height of the ball above the
ground?
b. What is the ball’s impact speed when it hits the
ground?
Draw velocity, acceleration and position graphs
Example Problem #2 – turning point
Bob throws the ball straight up at 20 m/s releasing the
ball 1.5 m above the ground.
a. What is the maximum height of the ball above the
ground?
b. What is the ball’s impact speed when it hits the
ground?
Step 1: pictorial representation, using particle model.
Pictorial Representation
– Draw an x or y axis and label the origin. Right/up are
positive directions. This is a position, not time axis.
– Put a dot at each important point in the problem, i.e. the
beginning, the end, each change of acceleration, turning
point (v=0).
– List the kinematic variables (position, time, velocity) at
each dot. At the first dot, x0, v0, t0, at the second dot, x1,
v1, t1, and so on.
– Show a0 between the 0 variables and the 1 variables with
an arrow showing the appropriate direction of
acceleration. Repeat as required.
Bob throws the ball straight up
at 20 m/s releasing the ball
1.5 m above the ground.
Y axis with the ground chosen
to be the origin (arbitrary
decision)
Three important points: ball
release (y0), high point (y1),
impact point (y2), which
happens to coincide with the
origin.
Acceleration due to gravity on
the way up and down so no
change in acceleration.
It’s not necessary to draw Bob!
2. List “knowns” and
quantities you need to
find.
Knowns
Find
y0 = 1.5m
y1, v2
v0 = 20 m/s
t0 = 0s (beginning of story)
a0 = -9.8m/s2
v1 = 0 m/s (high point)
y2 = 0 m (as we defined it!)
a. What is the maximum
height of the ball above
the ground?
b. What is the ball’s impact
speed when it hits the
ground?
3. Decide which equation to
use to solve part a. Does
the object have constant
v? No. Is time important
(a “known” or a find)?
No. Use the equation
without time:
v2 = v02 + 2a (∆y)
Knowns
y0 = 1.5m
v0 = 20 m/s
t0 = 0s
a0 = -9.8m/s2
v1 = 0 m/s
y2 = 0 m
Find
y1, v2
Bob throws the ball straight up at 20.0 m/s
releasing the ball 1.5 m above the
ground.
a. What is the maximum height of the
ball above the ground?
For part a, is time important? The
answer is no, so use equation #3 with
appropriate symbols:
v12 = v02 + 2a0∆y1,0 where ∆y1,0 = y1- y0
Solve for ∆y1,0 = - v02/2a (negative from
moving v02 to the left side of equation)
∆y1,0 = 20.4 m, so y1 = 21.9 m.
Knowns
y0 = 1.5m
v0 = 20 m/s
t0 = 0s
a0 = -9.8m/s2
v1 = 0 m/s
y2 = 0 m
Find
y1, v2
Bob throws the ball straight up at 20.0 m/s
releasing the ball 1.5 m above the
ground.
a. What is the maximum height of the
ball above the ground?
b. What is the ball’s impact speed when it
hits the ground?
For part b, is time important? The
answer is no, so use equation #3 with
appropriate symbols:
v2 = v1 + 2a0∆y2,1 where:
2
2
∆y2.1 = y2 - y1 = -y1
Solve for v2 = +/- (- 2a0y1)1/2
v2 = -20.7 m/s (it’s negative b/c ball is
going down
Assess: reasonable , since it fell just a tad
more than when it was launched.
Knowns
y0 = 1.5m
v0 = 20 m/s
t0 = 0s
a0 = -9.8m/s2
v1 = 0 m/s
y2 = 0 m
y1 = 21.9 m
Find
v2
How high does it go?
A coin is tossed into the air with an initial velocity of
8.0 m/s.
How high does it go before starting to fall?
Does it take the same amount of time to fall back to it’s
starting point as it took to go up? How can you show
this?
We usually use “y” for vertical motion
How high does it go?
http://bcs.wiley.com/hebcs/Books?action=mininav&bcsId=4678&itemId=04
70223553&assetId=160341&resourceId=15300
When different accelerations are involved, you
must divide the problem into parts – one for
each different acceleration
A “rocket sled” starts at rest and accelerates at 50 m/s/s
for 5.0 s. It runs out of fuel and coasts for 3.0 s, with
no change in speed. A braking parachute deploys and
the sled slows to a stop, accelerating at a rate of 3
m/s/s to do so.
A. What is the maximum velocity of the sled?
B. What is the total displacement of the sled?
Draw a pictorial representation for this problem with the
correct number of important points and list knowns
and “finds” with appropriate subscripts. Acceleration
should have correct signs (use diagram if not sure).
A “rocket sled” starts at rest and accelerates at 50 m/s/s for 5.0 s.
It runs out of fuel and coasts for 3.0 s, with no change in speed. A
braking parachute deploys and the sled slows to a stop,
accelerating at a rate of 3 m/s/s to do so. Find the max velocity and
the total displacement of the sled.
Note difference between t2 (8s after t0) and ∆t2 (t2 - t1), which is
3s.
v 1 = v2
0
Rocket sled problem
A. What is the maximum velocity of the sled?
0
Time is important. Maximum
velocity occurs in the first
segment, where sled speeds up:
vmax = v1
v1 = v0 + a0(t1-t0) = 250 m/s
v 1 = v2
Rocket Sled Problem
B. What is the total displacement of the sled?
Time is important for the first 2
segments. Find final position in
each segment with different
acceleration:
x1 = x0 + v0 (t1-t0) + ½ a0 (t1-t0)2
x1 = 625 m
x2 = x1 + v1 (t2-t1): constant v
x2= 1375m
No time is given for the 3rd
segment: v32 = v22 + 2a2(x3 - x2)
x3 = 11,800 m, which is the
total displacement.
v 1 = v2
Rocket sled problem, answers
• vmax = v1 = 250 m/s
• Δxtot = (x3 – x0) = x3 = 11,800 m
Different accelerations
A car starts from rest at a stop sign. It accelerates at 4.0
m/s/s for 6.0s, coasts for 2.0 s, and then slows down
at a rate of 3.0 m/s/s to come to a complete stop right
at the next stop sign. How far apart are the stop
signs?
• Draw a pictorial representation, choose a zero, list
knowns and “finds” and determine whether time is
important for each segment of the motion.
Different accelerations - answer
A car starts from rest at a stop sign. It accelerates at 4.0
m/s/s for 6.0s, coasts for 2.0 s, and then slows down
at a rate of 3.0 m/s/s to come to a complete stop right
at the next stop sign. How far apart are the stop
signs?
Answer: Δxtot = x3 = 216 m. Time was important in the
first segment, and we need to use both position and
velocity equations to get v1 and x1. The next segment
involved constant velocity, so there is only one
equation to choose. The 3rd segment did not involve
time so we could choose the “timeless” equation.
Free Fall Problem (#52)
• A pellet gun is fired straight downward from the edge
of a cliff that is 15 m above the ground. The pellet
strikes the ground with a speed of 27 m/s. How far
above the cliff would the pellet have gone if the gun
had been fired straight upward? (Note: the bullet has
the same initial speed whether fired up or down).
• For 2 object problems, there is always at least one
quantity that is the same for both objects. What is it for
gun shooting up and gun shooting down? There may
be more than one! Use your pictorial representation
and list of variables to figure out the important quantity
or quantities.
HardFree Fall Problem (#52)
v0D = -v0U = -20.9 m/s.
How far the bullet goes is y1U = 22.3m
2.6 Freely Falling Bodies
• If the only acceleration acting on an object is the
acceleration due to gravity, the object is said to be in freefall .
• Acceleration due to gravity is a vector with a magnitude
of 9.8 m/s/s and a direction towards the center of the earth
(“true vertical”). An object in freefall will change its
velocity by just under 10 meters per second every second if
it is traveling towards or away from the center of the earth.
For this chapter, that means up or down.
• The direction in which the gravity vector points is the
operational definition of “true vertical”.
• If on your paper, “down” is a negative direction (and it
often is!) the only non-zero component of gravitational
acceleration in negative.
• Free fall refers to objects going up or down, with an
initial velocity or not, if the only acceleration is due to
gravity.
Kinematic Equations for Constant
Acceleration
• x = x0 + vconst (∆t) (equation of a line for motion with
constant velocity; this is the only equation of motion
if a = 0).
• v = v0 + a(∆t) (equation of a line for motion with
constant acceleration; use if time is important).
• x = x0 + v0 (∆t) + ½ a (∆t)2 (quadratic equation, use if
time is important)
• v2 = v02 + 2a (∆x) (Use this equation if time is not
important).
• ∆x = ½(v0 + v) (∆t) ( where ½(v0 + v) is vavg).
All velocities are instantaneous unless noted
Analysis of a position graph
In which segment does the
velocity have the greatest
value?
A.
B.
C.
D. Not enough information to
determine.
Analysis of a position graph
In which segment does the
velocity have the greatest
value?
A.
B.
C.
D. Not enough information to
determine.
Analysis of a position graph
In which segment is the motion
in a negative direction?
A.
B.
C.
D. Both B and C
Making a velocity graph from a position
graph
Draw a numerically
correct velocity-time
graph for this positiontime graph.
Motion graph for constant acceleration
http://bcs.wiley.com/hebcs/Books?action=mininav&bcsId=4678&itemId=04
70223553&assetId=160341&resourceId=15300
Predict the graphs for position vs. time and velocity vs
time for the spaceship.