### Present Remotely

Send the link below via email or IM

• Invited audience members will follow you as you navigate and present
• People invited to a presentation do not need a Prezi account
• This link expires 10 minutes after you close the presentation

Do you really want to delete this prezi?

Neither you, nor the coeditors you shared it with will be able to recover it again.

# Kinematic of motion

Chapter 2-Kinematic concepts: the physics of objects in motion
by

## Keith Herold

on 2 October 2016

Report abuse

#### Transcript of Kinematic of motion

The physics of objects in motion
Displacement & velocity
KINEMATIC CONCEPTS
Displacement "s"
velocity (a vector quantity) "v"
-should really have a vector above the v...
speed can be a "v" also but without direction.
s=distance/displacement
Some fundamental and key terms to know before we venture too far.
displacement can be in a positive or negative direction. Total displacement is the actual distance you have moved from your point of reference. *=delta or change {prezi doesn't have too many symbols so the asterik * will = delta or "change in"
*x = final displacement - initial displacement
x
Sample example:
An object moves from reference pt. 0 to the left 12 meters. The object then moves 13 meters to the right. What is the objects total displacement?
Solution: -12+13=*x
*x=1
Sample example ii.
An object has a displacement of 7m. It moves a distance to the left equal to 15m and then 22m to the right. (a)What is the total distance traveled? (b)What is the change in displacement?
Solution:
a. 15+22=37m total distance traveled
b. *x=final x-initial x
*x=[(7+(-15)+22]-7
-8+22-7
7m
Speed=distance traveled per time
v=s/t
The velocity would incorporate

Speed and velocity are really instantaneous concepts. In other words,,,,at any particular time the speed or velocity is a given amount but in a second or minute, it may be quite different...Therefore we are calculating the average speed or velocity generally.
e.g.: If an object travels a total distance of 682.5 km in a four hour period, what is the average speed of that object given in meters/second?
Solution:
first change the kilometers to meters by simply mux. by 1000.
682.5x10^3 in stand. form Sci. not.
v=s/t v=6.825x10^3 ; time must be converted to seconds. Nb: there are 60min.x 60sec. or 3600 sec. in one hour.

Therefore 4x3600=14,400sec or 1.44X10^4
v=6.825x10^5/1.44x10^4
= 47.4 m/s
We can find the speed at any particular exact moment in time by using the equation: aka instantaneous speed
v=*s/*t ...........Nb: by definition speed (not vel.) is always positive!

v=u+at vel=(init. vel) + (accel.)(time)

s=ut+ 1/2 at^2 displace.= init. vel. x time +[ 1/2 accel. x time squared]
Ex. prob.:
The initial displacem't of a body moving a constant velocity of 35 m/s is = -45 m. (a).When does the body reach the point of displacem't of 10 m? (b) What dist. does the body cover in this time?
---write down the given:
speed=35 m/s
Xo =initial displacem't=(-45m)
Total displacem't X=10 m

Solution:
X=Xo+vt
10=(-45)+35t
(a) t=1.57 sec
*since you have time now...put back into the original equation that V=distance/time
(b) X=vt \
X=35 (1.57)
X=55 meters traveled.
Problem: Q15 Tsokos, page 45:
A mass starts out from O with a vel. of 10m/s & continues moving at this vel. for 5 s. The vel. is abruptly reversed to (-5 m/s) & the object moves at this vel. for 10s. Find

a. change in displacement
b. tot. dist. traveled.
c. aver. speed
d. aver. vel.
Solution: (a) since V=s/t; part one of journey:
s=vt =10m/s(5s)=50m.

part 2 of journey: s=(-5m/s)10s=-50m
50+(-50)=0 displacement [back at starting point.

(b)100 meters is total dist. traveled [50+50]

(c) the aver. speed is total dist. traveled/t=100m/15s
=6.67 m/s
(d). Aver Vel is total displacement traveled/time
=0m/15sec=0......
FRAMES OF REFERENCE:
observer A on the ground
observer B-a passenger on a train
observer C-who is on the train walking.
Each observer describes their velocity with respect to their frame of reference or position. So therefore it is an important point to note in problems/situation what is the reference point!!!!! (otherwise your results/solutions will be different....
If a tomato is tossed vertically upwards if will land exactly where the reference point is, the hand...
What about if a tomato is tossed upwards (vertically) from a convertible car going 100 km/hour, where will it land????
The frame of reference in the tomato/convertible case is again the hand of the tosser.
Realize their is a horizontal component to the car and the same for the tomato.
What is uniform motion?
well.....it does mean the graph of the motion will be a straight line if we plot velocity vs. time...
Recall: velocity is the displacement or distance/time
If there is not change in the speed and direction this is uniform motion.
The change in velocity per time is called acceleration. No change (speed or direction)=no acceleration!
S
o if a body goes for 2.00 hours and makes a distance covered of 150. km north, what is the average velocity?

V=dist. / time
150km x 1000m/(2x60 min. x 60 sec.)
V= in m/sec !!!
V=2.0833333x10^4
V= 2.08 x 10^4 m/s
***Nb: the three significant digits in both and the answer in 3 sig. figs.!
This is the actual accepted standard of how to do answers in science....
ACCELERATION!
acceleration is the change in vel. pre change in time.
a=*v/*t
Motion with constant acceleration:
-if the accel. is not changing. *see graph of Vel. vs time=accel.
the motion is uniform and constant. This also takes into account there is no change in direction.
Concept of instantaneous acceleration:
-the point a a particular instant in time or the tangental line to a graph of acceleration vs. time.
For motion in a straight line: positive velocity is in a + direction and negative in a - direction.
But for acceleration + means increasing accel. and neg - is decreasing acceleration

ACCELERATION DUE TO GRAVITY
OFTEN a=g
-9.81 m/s^2 -because it is going into a down direction. We often dismiss the negative.
V^2=U^2 + 2as
An equation to find the final velocity at some point in time. This can be with free fall due to gravity (since you know the g-value=a=9.81 m/s^2) or other problems.
Data Booklet equations:

s: displacement
t: time
u: initial speed
v: final speed
a: acceleration
Lets do a problem:
0.6 m/s
0.8 m/s
The diagram shows a boat that is about to cross a river in a direction perpendicular to the bank at a speed indicated(0.8m/s), also see the current speed(0.6 m/s).
What is the magnitude and displacement of the boat 5 seconds after leaving the bank of the river?
Here's where a vector diagram can really help solve the problem...
In 5 seconds the boat will have an upward displacement of 4 m. The water current will give the boat a displacem't of 3m tothe right. Draw the vectors....the final answer is the resultant!=5m NE.
A sprinter leaves the starting blocks of the 100m dash race and accelerates to a top speed of 11.8 m/s in 2.6 s. Calculate the average acceleration of the sprinter in the first 2.6 seconds.
accel. (aver.)= *v/*t
a=11.8-0/2.6-0
a=4.5 m/s^2
When objects get dropped or fall, they do so in a negative direction. So it is conventional to show downward displacement as negative. Also downward vel's and accel's vectors should also be shown as neg.
To figure the displacement after so many seconds you need to use the equation: s= ut + 1/2 at^2
What is the displacement of an object after 1 second?
s= (0)x1 + 1/2(-9.8)x 1squared
s=4.9meters
What would the displacement be after 5 seconds?
s=1/2(-9.8)(5^2)
s=-4.9x25
s=-122.5 m
Jason hits a volleyball so that it moves with an initial velocity of 6.0 m/s straight upward. If the volleyball starts from 2.0m above the ground how long will it be in the air?[before striking the earth].
Let's approach the problem by changing the velocity and time information into displacement

Since velocity = dist/time or s/t then
dist s=(v)t
0.6(5)=3. & 0.8(5)=4.
y=4 and x=3 ; since it makes a 3-4-5 triangle the hypotenuse is 5.
The instantaneous velocity is the tangent to the curve (i.e. the slope) at any given point
Given: u=6.0 m/s & 2 meter off ground in launch.

If I determine the distance up, then I can determine the time up as well.

v^2=u^2 + 2as
0^2=6^2 + 2(9.81 m/s)s s= 1.8 meters up

*nb: you could use v=u +at finding time first...
v=u+at 0=6m/s + 9.81m/s (t) t=.6116 seconds
plug into: s= (u+v)/2 x t s=(6+0)/2 x .6116= 1.83 m up.
Add 1.83 meters and 2 meters= 3.83 meters to fall
s=ut + 1/2 at^2 u=0 at start of fall so 1/2at^2
rearrange equation to time t=square root of (2s)/a
t= `0.8836sec down plus the time up at .6116
total time = approx. 1.5 seconds total flight.
Full transcript