with Gym :
Aflevering 2 fysik og mekanik
44.
a)
I set the time the bus leaves the stop to be ti = 0. I set the bus stop at xi = 0 and
positive x direction in the direction they both are going.
This gives me some extra values beside the ones the problem gives me. I will
list the known values for the bus and the person
The bus:
vi, bus = 0
abus = 2.0
m
s2
The person:
m
vp = 6.0
s
ap = 0
xi = 8.0 m
I know that the bus and person has to arrive at the same x position at a given
time, so xbus = xp , which gives me an equation i can use
m
1
m
xp = 8.0 m 6.0
t
0 t2 = 8.0 m 6.0
t
s
2
s
1
m
1
m
xbus = 0 0 t
2.00 2 t2 =
2 2 t2
2
2
s
s
Because i know the x positions has to be equal at the time she reaches the
bus, i can set the two equations equal an solve for t
1
solve 8.0 6.0 t =
2.0 t2, t
2
2., 4.
So the first time she reaches the bus is after 2.0 s
(1)
b)
Because her acceleration is 0 her speed is still 6.0
the bus
m
when she reaches
s
c)
i know the initial velocity of the bus, how long the bus has travelled when she
catches up and the acceleration of the bus, so i have
v t =0
2.0
m
2s
s2
4.0 m
s
v t =
(2)
So the velocity of the bus when she draws even is 4.0
d)
v(t) for the bus is
m
v t = 2.0 2 t
s
It is plotted below
plot 2.0 t
20
10
10
0
5
5
10
t
10
20
v(t) for the person is
m
v t = 6.0
s
It is plotted below
plot 6.0
9
8
7
6
5
4
10
e)
5
0
5
10
m
s
x(t) for the bus is
1
m
x t =
2.0 2 t2
2
s
It is plotted below
1
plot
2.0 t2, t = 0 ..10
2
100
80
60
40
20
0
2
4
6
t
8
10
x(t) for the person is
m
x t = 8.0 m 6.0
t
s
It is plotted below
plot 8.0 6.0 t, t = 0 ..10
50
40
30
20
10
0
2
4
t
6
8
10
58.
a)
I list the values i know
m
vballoon = 12
s
m
acamera = g = 9.82 2
s
xi = 18 m
m
vi, camera = 12
s
xf = 0
With these values i can first calculate how long it takes the camera to reach the
ground with the solve function
solve 0 = 18
12 t
1
9.82 t2, t
2
3.493396777,
1.049404924
(3)
Given that time cannot be negative it takes the camera 3.49 s to reach the
ground. I can now calculate how far the hot-air balloon rises in that amount of
time.
xf = 18 m
12
m
3.493396777 s
s
xf = 59.92076132 m
(4)
So the railing is 60 m above the ground when the camera hits the ground
82.
a)
If we insert the given values in the expression for v and make it a function of
time, we get
m
v t = 0.040
s
1 e
t
0.50 s
Since a(t) is the derivative of the velocity with respect to t, we get a(t) by
deriving v(t).
First i simplify v(t)
v t = 0.040 1 e
t
0.50
= 0.040
So i have to derive
dv
0.040
dt
First i use the difference rule
0.040e
2.0 t
0.040e
2.0 t
f g ' = f' g' = 0.040 '
0.040e
2.0 t
'= 0
0.040e
2.0 t
'
1
For the g' part i factor out the constant
= 0.040
dv
e
dt
2.0 t
And then i use the rule of composite functions and first derive the functions
dv t
e = et
dt
y:
dv
dt
i :
2.0 t = 2.0
then i compute
0.040e
2.0 t
' = y' i
i '= e
2.0 t
2.0 = 2.0e
2t
I will insert that in (1) and also factor in the constant from earlier and get an
expression for a(t)
a t = v' t = 0
0.040 2.0e
2t
So the expression is a t = 0.08
b)
First i define the functions
v t
a t
0.040 1 e
0.08e 2 t :
then i plot v(t)
plot v t , t = 0 ..2
t
0.50
:
= 0.08e
m
e
s
2t
2s t
0.03
0.02
0.01
0
0.5
1
t
1.5
2
And a(t)
plot a t , t = 0 ..2
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
0.5
1
t
1.5
2
c)
Because v(t) is the derivative of x(t), then v(t) integrated will give x(t). I use the
integrate function
int 0.040 1 e
t
0.50
,t
= 0.04000000000 t
0.02000000000 e
I will use the integration test to check the result
diff 0.04000000000 t
0.04000000000
Which is exactly v(t).
So an expression for x(t) is
x t
0.04 t
0.02e
I plot the graph
plot x t , t = 0 ..3.5
2.0 t
:
0.02000000000 e
0.04000000000 e
2. t
,t =
2. t
2. t
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0
1
t
2
3
d)
The sphere was dropped at the surface which is at x = 0.02 m so the bottom
will be at x = 0.12 m.
I read the graph and find that i takes about 3 s for the sphere to reach the
bottom of the container