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A) For A, I interpreted the
question as to find the time
when the trooper overtakes
the car, meaning find when
the final position of the
trooper and the driver are
equal. Using both the
kinematics equations, I
wrote both the driver’s and
trooper’s equation in terms
of X final, in order to solve
for t.
B) Using what we
solved, we know that the
trooper overtakes the car
at 31.88 seconds. We
can use this to find to
the final velocity of the
trooper using
kinematics, since we
know the trooper starts
accelerating from rest,
meaning initial velocity is
C) Using what we solved for and our
given, we can find our delta x /
distance the trooper traveled to
overtake the car using the kinematic
equations knowing our final velocity of
the trooper and the time it took to
overtake the driver.
D
:
Driver
given
:
T
:
Trooper
V
=
Vp
= 45m/s (Constant speed)
at
=
3
M/se
stats
cycle
Di
ap
=
0m/s
accelerates
tp
=
(Prons)
Voc
= Om/s (from
rest)
ahead
=
a) t
when
X
b)
Vi
,
at t
. 88s/
,
when
Xt
i
fi
fi
Use X-Xo
=
Vit-tat use VE
,
V
t
e
Driver
:
X
1 He
Vipt
,
FD/in
VF
=
(3)(
.
Trooper
:
X-X
a
VF
= 95
.
64M/s
X
fr
Ift
C)
Ax
using
our
given
:
VFF9SG
e
2
Viptp
=
a
t Ax
:
Vottt
Vip(t
=
a
t rest
Vipt
2Vip
=
54
t Ax
=
+(3)(
.
=
Ea
-Vist-ZVid
Ax
= 1524
.
50m
= E (3)
45t
2(45)
t
.
88s
A) Using the given and breaking the initial
velocity into it’s components, we can use
the kinematics in terms of the y axis, as
we know the acceleration of y will be our
gravity, our initial velocity broken up into
its y component, and height / delta Y. We
can write this into a quadratic formula to
solve for t.
B) Since we know the stone is
thrown downward, its max
height it reaches will be the
height it starts off at, which is
45m.
C) To find our distance / delta x, we can use the
kinematics in terms of the x axis, as know our stone is
not accelerating in the x direction, our time, from
problem A, and the initial velocity’s x component.
D) In order to find the stone’s final velocity at the given
level, we must understand that we will have a new time in
which it will hit that specific level. We can’t use the 2.
seconds as that is our time for when the stone strikes the
ground. We can solve our time, by substituting our new
delta Y into the kinematics in the y axis, knowing our delta
Y as the gravity, and our initial velocity’s Y component. We
can then put it into a quadratic formula, to solve for t.
Since we solved for t, we must get its final velocity’s X and
Y components. We can do this using the corresponding
kinematics in the x and y axis, and also using our given.
Once we solved for both x and y components of the final
velocity, we can take the magnitude to find the final
velocity of the stone 15m level below the level of launch.
--
+, given
:
Voy
= 20 m/s'
,
300
ay
=
=
.
8 M/s
(
20cos(
%
)
45M
Voy
=
20 m/s'
,
20 m/s
ax
=
0m/s
20sin(-
%
)
Yi
= 45n
Y
=
0n3xy
=
45m
a)
t
for stone d) VF
When
TSM
BAy
=
15 m
Use Y -Yo
:
Voytt'ayt
don't use
- 177s
,
we
need
= 20 sin +
%
.
to
find
our
new t,
given
our
=
E
20 sin ( 30
+ 45m
new
Ay
=
.
177s Use Y - Yo-
Voytt
tayt
b)
Max
height
Ay
=
Voyttt
myt
15m
=
Gayt +opt
Max
height
would
=
6
8n/s") t
(
sin
+ /Sm
equal
initial height
since
it was
thrown
t
=
005s
downward
.
find
Up
which
is
(Ex)"
(VFy)
= 45
.
Om
Vi
=
Voxt
aft
C) AX
when +
= 2 .
177s
VFx
= 20cos(
%
Use Ax
=
Voxt
! a
e
=
32 n/s
ACCELERATING
Viy
:
Voytayt
Ax
=
(20n(s(os
(
%
)(2.
177s)
Vfy
=
20 sin
(
8n(sY)(
. 005s)
=
71m
VFy
=
19 .
85m/s
VF
:
.
VF
=
.
m/s
A) To find our resultant force, we must add up all our
forces, taking into account each of our components. I
divided the problem into 4 graphs, because it is 4
vectors and I written each vector with its components
depending on the direction and angle, as we defined
the 1st quadrant as our positive x axis. Then can write
our final resultant force vector.
B) To find our magnitude, we have the take the square
root of our components squared, in which we get
199.47m.
To find our angle, we have to find theta from our
tangent, in which we originally get 51.64 degrees. I add
180 degrees, as it asked to use the positive x axis as
reference, and because the resultant vector is pointed
towards quadrant 3.
·
&
·
·
M
a)
R
=?
R
=
F
B
F
xY
B
xY
E xY
D ↑
%
2100
from
1180
From
00 + X
uxis
100M
&
10
=
Y -
X
= - X =
30
°
= - X
258r
(
(
(
ISOcos(210%)
,
150sin (
%
) (200cos(18%) ,
200 sin (1180)
,
0 <
,
,
- 89
,
176 .
R
= <
40
,
.
E
=
<
,
.
4107(m)
as
F
=
.
b)
IRI and
O
(R
=
H
.798)"+ ( .
IR
=
47 M
O
=
tan"
(
=
tan"
(408)
. 6
:
①
= 51 .
64
°
O
=
. 640 from
A) To find acceleration total, we must understand
that the total acceleration can break broken up into
individual components, just like x and y components.
To find our acceleration tangential, we can use our
kinematics as we our given our final velocity and
initial velocity. First we must convert km/hr to m/s by
using dimensional analysis. Then we can find our
tangential acceleration by solving for a. We get -1.
m/s squared.
To find our centripetal acceleration, it says to find it
at 60 km/hr, which is our final velocity. Use the
equation for centripetal acceleration, and our radius
which is given to us. We get 1.85 m/s squared.
Finally, we can find our total acceleration which is the
magnitude of both the centripetal acceleration and
tangential acceleration. We get 2.157 m/s squared.
W
f
given
:
100
kn/hr
>
Gokm/hr
Vi
=
lookn
,
1000 3600s
slows
G
Ik(z)
down
X
78m/s
16 .
67 M/s
60km
Ik
V
=
= -
1000
G
kn
~ 3600s
(
=
)
t
= (Os
r
=
ISOR
a)
=
find &TOTAL
A
Total
(
) + (a
,
)
A
:
Acceleratio
one
use Ve
=
Vo
a
t
a
= V
t
a
=
67M/s
78m/s
los
=
H
((n/s
A
=
Centipati
a
act
6
.
85n/s
A
TOTAL"
(-.
llm/sY"
(
.
85
m/sY)
A
Total
=
- 157 m/s
