Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
These are the fundamental points in the following Lecture Slides : Motion Problems, Solving Motion Problems, Discrete Case, Continuous Case, Constant acceleration, Horizontal, Motion, Vertical, Discrete Case, initial Position
Typology: Slides
1 / 26
This page cannot be seen from the preview
Don't miss anything!
In solving motion problems, you should always
remember to start from the basics.For the
discrete case
, these are:
v
x-average
x
t^
and
a
x-average
v
x
t^
..
For the continuous case when we have constant acceleration
, these become:
v(t) = v
o^
and
s(t) = s
o^
t + ½ ato^
2
(where s can be either x or y
depending on whether the motion is horizontal orvertical).
Given the following values of velocity at the
indicated times, find the average accelerationbetween t = 4 seconds and t = 5 seconds, andfind the position at t = 5 seconds given that theinitial position is -7 m.t (s)
v (m/s)
t (s)
v (m/s)
Since we are working with discrete values, we start
with the definitions of average velocity andaverage acceleration: a
x-average
v
x
t^
and
v
x-average
x
t
For the average acceleration between t = 4 s and t
= 5 s, we can apply the average accelerationformula directly: a
x-average
v
x
t^
becomes
a
avg
v
x
t = (7.95 m/s – 10 m/s) / (5s – 4s)
= 2.05 m/s
2
o^
x-average
or
x = x
f^
- x
i^
= v(between t
f^
and t
)(ti*
-tf^
)i
x
4
= -3.98 m + ½(7.95 m/s + 10.0 m/s) * (4s – 3s)= -3.98 m + (+8.98 m/s)(1 s) = 5.00 m = x*
4
x
5
5.00 m + ½(10.0 m/s + 7.95 m/s) * (5s – 4s) = 5.00 m + (+8.98 m) = +13.98 m = x
5
In this particular case, the velocity data was obtained from the
functional form:
v(t) = +3 m/s – (7 m/s)*cos({
rad/4 sec}
t)
[note that {
^
rad / 4 sec} = {
o /sec} ]
Which leads to functional forms for x(t) and a(t) of:
a(t) = (7*
m/ 4 s
2 )*sin({
rad/4 sec}
t)
x(t) = -7 m + (3 m/s)*t – (28 m/
sin({
rad/4 sec}
t)
a(t) = (7*
m/ 4 s
2 )*sin({
rad/4 sec}
t)
x(t) = -7 m + (3 m/s)*t – (28 m/
sin({
rad/4 sec}
t)
If we use the above functional forms to calculate the
acceleration between t = 4 sec and t = 5 sec, that is, at t= 4.5 sec, we get: a(t) = (7*
m/ 4 s
2 )*sin({
rad/4 sec}
4.5 sec) = -2.10 m/s
2
which compares to our numerical result of -2.05 m/s
If we use the above functional forms to calculate the
position at t = 5 sec, we get:
x(5 sec) =
-7 m + (3 m/s)*(5 s) – (28 m/
sin({
rad/4 sec}
(5 s)) =
-7 m + 15 m + 6.30 m = 14.30 m which compares to our
numerical result of 13.98 m.
A plane takes off from an aircraft carrier. Assume
the following:
-^
the plane accelerates with a constantacceleration,
-^
the flight deck is 80 meters long,
-^
the initial speed of the plane is zero,
-^
the final speed of the plane is 85 m/s. What is the acceleration of the plane during take-
off, and how long a time does the take-off take?
x = 80 meters
v^ o
= 0 m/s
vf^
= 85 m/s
to^
= 0 s
tf^
=?
a =?
o^
o^
2
2
2
2
Are these results reasonable?A plane taking off from an aircraft carrier on a short flight
deck certainly has to take off in a short time, so about 2seconds looks reasonable. The acceleration of 45.16 m/s
2
is equivalent to 4.61 gees’
(1 gee = 9.8 m/s
2 ). This is a pretty big acceleration, but
that also sounds reasonable given the circumstances.By the way, this type of acceleration will tend to makemost people black out – not a good thing when taking offin a plane.
v(t) = v
o^
+ at*
and
y(t) = y
o^
+ v
t + ½ ato
2
where we identify the following:y = 15 m
(final position is 15 meters above the ground)
y
o^
= 0 m
(initial position is on the ground)
v =?
(speed when it reaches the person on the balcony)
v
o^
= 3 m/s
(initial speed
t =?
(time when it reaches the person on the balcony)
a = g = -9.8 m/s
2
(acceleration due to gravity)
o^
o^
2
2
