Transcript Chapter 2 Examples

Chapter 2 Examples
Problem 2.46
An egg is thrown vertically upward from a point near the
cornice of a tall building. It just misses the cornice
on the way down and passes a point 50 m below its
starting point 5.0 s after it leaves the thrower’s hand.
Ignore air resistance.
a.
What is the initial speed of the egg?
b.
How high does it rise above its starting point?
c.
What is the magnitude of its velocity at its highest
point?
d.
What are the magnitude and direction of its
acceleration at the highest point?
Step 1: Draw it
h, height above building
50 m in 5 s
Make some definitions

Let t=5 s
 Let vo= initial
velocity, in positive
direction
 Let g=9.8 m/s2 ,
acceleration due to
gravity, in negative
direction.
 total distance =-50 m
v(t )  at  vo
1 2
y (t )  at  vot  yo
2
v2f  v02  2a( y f  yo )
Apply Definitions and Solve




Let t=5 s
Let vo= initial velocity, in
positive direction
Let g=9.8 m/s2 ,
acceleration due to
gravity, in negative
direction. This will be
“a”
total distance =-50 m
=y(t)-yo
1 2
y (t )  at  vot  yo
2
1 2
y (t )  yo  at  vot
2
1
50   g (5) 2  vo (5)
2
122.5
72.5  5v0
14.5 m / s  v0
Part B) Find height above bldg
 v0=14.5 m/s
 At top of the
arc, vf=0
 At t=0, let y0=0
v(t )  at  vo
1 2
y (t )  at  vot  yo
2
v2f  v02  2a( y f  yo )
Part B) Find height above bldg

v0=14.5 m/s
 At top of the arc, vf=0
 At t=0, let y0=0
 Let yf=h
v2f  v02  2a( y f  yo )
v 2f  v02  2 g (h  0)
0  14.5 
2
2 g
10.7 m  h
h
Parts C) and D)
At the top of the arc, v=0 m/s
 Always in this problem a=-g=-9.8 m/s2

Problem 2.61
A car 3.5 m in length and traveling at a constant
speed of 20 m/s is approaching an intersection
which is 20 m wide. The light turns yellow
when the front of the car is 50 m from the
beginning of the intersection. If the driver
steps on the brake, the car will slow at -3.8
m/s2. If the driver steps on the gas pedal, the
car will accelerate at 2.3 m/s2. The light will be
yellow for 3 seconds. Ignore reaction time. To
avoid being in the intersection when the light
changes to red, accelerate or brake?
Step 1: Draw It!
3.5 m
In order for the car to
run the green light,
It must traverse
3.5+50+20 m=73.5 m
so that no part of the
car is in the
intersection
50 m
20 m
Braking is easier, it
must traverse less
than 50 m
Our Options

We know that
t=3 s
 a is either 2.3
m/s2 or -3.8
m/s2
 vo=20 m/s
 Total distance is
either 73.5 m or
50 m

v(t )  at  vo
1 2
x(t )  at  vot  xo
2
v2f  v02  2a( x f  xo )
Our Options

1 2
x(t )  at  vot  xo
2
We know that
 t=3 s
 a is either 2.3 m/s2 or -3.8 m/s2
 vo=20 m/s
 Total distance is either >73.5 m or <50 m
Braking
1 2
at  vot
2
1
x(t )  xo   (3.8)  32  20  3
2
x(t )  xo  42.9 m
x(t ) 
Run It !
1 2
x(t )  at  vot  xo
2
1
x(t )  xo  (2.3)  32  20  3
2
x(t )  xo  70.4 m
Problem 2.77
A physics student with too much free time
drops a water melon from the roof of a
building. He hears the sound of the
watermelon going “splat” 2.5 s later.
How high is the building? The speed of
sound is 340 m/s and ignore air
resistance.
Step 1: Draw It!
h
Splat!
Some Hard Thinkin’
The melon experiences an acceleration
due to gravity. The student merely
dropped it, so its initial velocity was 0.
 The sound wave is unaffected by gravity
so it moves with constant velocity from
the ground toward the student.
 These are two separate events with a
total time of 2.5 s

Some Hard Thinkin’ part 2
The equation for the distance that the
melon traverses is y=-1/2*g*(t1)2 where
y= height of bldg and t1 is the time for the
fall.
 The equation of distance for the sound
wave is y=vs*t2 where vs = speed of
sound =340 m/s
 The total time for all this to transpire is
2.5 s or 2.5 s =t1+t2

Solving
1
 y   gt12
2
y  340t2
 y  4.9t12
2.5  t1  t2
 t2  2.5  t1
y  340*  2.5  t1   850  340t1
4.9t12  850  340t1 or 4.9t12  340t1  850  0
b  b 2  4ac 340  3402  4  4.9  (850)
t1 

2a
2  4.9
t1  2.416 or 2.42 s
y  4.9* 2.422  28.6 m
or
t2  2.5  2.416  0.084
y  340*0.084  28.6 m
Problem 2.89
A painter is standing on scaffolding that is raised at a
constant speed. As he travels upward, he
accidentally nudges a paint can off the scaffolding
and it falls 15 m to the ground. You are watching and
measure with your stopwatch that it takes 3.25 s for
the can to reach the ground.
a. What is the speed of the can just before it hits the
ground?
b. Another painter is standing on a ledge with his hands
4 m above the can when it falls. He has lightningfast reflexes and can catch if at all possible. Does he
get a chance?
Part a) Make some definitions

Let t=3.25 s
 Let vo= initial
velocity, in positive
direction
 Let g=9.8 m/s2 ,
acceleration due to
gravity, in negative
direction.
 total distance =-15 m
v(t )  at  vo
1 2
y (t )  at  vot  yo
2
v2f  v02  2a( y f  yo )
Make some definitions

We must solve this
equation for vo
And
then we must
use this equation to
solve for the
velocity
v(t )  at  vo
1 2
y (t )  at  vot  yo
2
v2f  v02  2a( y f  yo )
Solving
1 2
y (t )  at  vot  yo
2
y (t )  yo  15
1
15   g (3.25) 2  vo  3.25
2
36.75  vo  3.25
11.31 m / s  vo
v(t )  at  vo
v(t )   g *3.25  11.31
v(t )  20.5 m
Part B) Make some definitions






The word “falls” is slightly
misleading, the can first rises
in the air and then falls to the
ground.
What it is asking for is how
high the can flies above the
release point; it must be
greater than 4 m
So we are solving for a total
“y” displacement (yf-yo)
At the top of the arc, vf=0
Let vo= initial velocity, in
positive direction, 11.31 m/s
Let g=9.8 m/s2 , acceleration
due to gravity, in negative
direction.
v(t )  at  vo
1 2
y (t )  at  vot  yo
2
v2f  v02  2a( y f  yo )
Part B) Make some definitions





What it is asking for is how
high the can flies above the
release point; it must be
greater than 4 m
So we are solving for a total
“y” displacement (yf-yo)
At the top of the arc, vf=0
Let vo= initial velocity, in
positive direction, 11.31 m/s
Let g=9.8 m/s2 , acceleration
due to gravity, in negative
direction.
v2f  v02  2a( y f  yo )
0  11.312  2 g ( y f  yo )
11.312
 ( y f  yo )
2 g
6.52 m  ( y f  yo )
Yes, he can
catch it