Kinematics In One Dimension - Physics - Free Educational videos for Students in K-12

Kinematics In One Dimension - Physics - By The Organic Chemistry Tutor

Transcript


00:01	in this video , we're going to talk about schematics
00:04	, which basically describes how objects move without any references
00:09	to force . Now we're going to focus on schematics
00:13	in one dimension , mostly along the X axis ,
00:18	but we can also work on some problems along the
00:20	Y axis . When you get into two dimensional mathematics
00:24	, it covers project emotion , which is a topic
00:28	for another day . Now the first thing we need
00:31	to talk about is the difference between a scale of
00:35	quantity and a vector quantity . A scale of quantity
00:42	is something that has magnitude only , whereas a vector
00:46	quantity has both magnitude and direction . So for instance
00:53	, mass is a scalar quantity . You could say
01:00	that a book has two kg of mass And the
01:04	two kg Would be the magnitude of the mass of
01:08	the book . But you wouldn't say the book has
01:11	two kg of mass . East direction wouldn't apply for
01:16	mass . So mass cannot be a vector quantity .
01:23	Distance is a scale of quantity and displacement is a
01:31	vector quantity , displacement . You can think of it
01:36	as distance with direction . So let's say if a
01:41	person traveled 15 m , you're basically describing a person's
01:46	distance because you didn't mention direction . But let's say
01:50	if someone traveled 15 m east now you're describing displacement
01:55	. You describe in not only how far he traveled
01:59	, but also where he traveled . So think of
02:01	displacement as distance with direction . Now technically displacement is
02:07	really the change in the position of an object .
02:12	So along the X axis you could describe displacement as
02:15	the final position , modest the initial position , speed
02:23	is a scale of quantity . Speed describes how fast
02:27	an object is moving . Velocity is a vector quantity
02:33	, velocity is speed with direction . So if you
02:37	were to say , a person is traveling at uh
02:41	20 m/s , you're describing the speed of the person
02:44	, this is the magnitude . But let's say if
02:47	a person is moving at 30 m/s north , you're
02:52	describing velocity because you've mentioned his speed along with the
02:58	direction . So you have both magnitude and direction ,
03:01	which makes it a vector quantity . Now , what
03:05	about temperature ? Would you say that temperature is a
03:10	scale of quantity ? Or would you describe it as
03:12	a vector quantity ? So , think of units of
03:18	temperature . Let's say it's a 80°F outside Or 25°C.
03:27	. You wouldn't say it's 80°F direction doesn't apply for
03:32	temperature . So because you can't describe the direction of
03:36	temperature , it's not relevant . Temperature is a scale
03:40	equality . You can only mention its magnitude , such
03:46	as how hot it is , 80°F. . But you
03:50	can't attach direction to temperature , which makes it scaler
03:54	now acceleration and that is a vector quantity , acceleration
04:04	tells you how fast the velocity changes . And you
04:08	could say a car is accelerating at two m per
04:12	second squared east , you can put direction to it
04:16	. So acceleration is effective quantity . Now let's talk
04:21	more about the difference between distance and displacement . So
04:29	let's say a person walk , Let's say a person
04:34	traveled 13 m east and then he turns around And
04:41	travels four m west . We're going to add a
04:46	negative science this because he's going towards the left and
04:51	just to review direction north and south is along the
04:54	Y axis , east and west there along the X
05:00	axis . So peace is along the positive X axis
05:05	. West is along the negative X axis . So
05:10	given a situation , calculate the distance and the displacement
05:15	of this individual that traveled here . Let's put a
05:20	person . So the total distance that this person traveled
05:32	is 17 m . Mhm . It's basically the some
05:38	of these two numbers he traveled 13 m east and
05:41	then four m west . So a total distance of
05:44	17 m distance being scalar , it's always positive .
05:52	Now there is an exception to this , specifically temperature
05:56	depending on the units you're dealing with . Even though
06:00	temperatures , the scale of quantity , you could have
06:01	negative values , especially for units such as Fahrenheit and
06:06	Celsius . So some places can be negative 20°F in
06:12	temperature , Others negative 10°C. . But on the kelvin
06:17	scale there's no negative values . So there are some
06:21	scalar quantities that do have negative values like temperature .
06:24	So just be aware of that . Now , going
06:28	back to this problem , now that we know the
06:30	total distance traveled , what is the displacement of this
06:35	particular individual ? Now , displacement can be positive or
06:41	negative dependent on the direction . If you were to
06:46	combine these two , including the negative four value ,
06:49	It would be 13 -4 , So the displacement is
06:53	positive . nine , displacement is the final position minus
06:59	the initial position . So displacement is the change in
07:03	position . What this tells us is that This person
07:08	ended up nine m east from where he started .
07:12	And so let's illustrate this with a number line .
07:16	So let's say that He started at the origin position
07:20	zero . He traveled 13 m east . So because
07:26	he's moving along the positive X axis , we assign
07:29	a positive value here . And then during the second
07:33	part of his trip He traveled four m west .
07:38	So 13 -4 . He ends up at position nine
07:44	. So the net result is that he traveled nine
07:48	m east and so that's his displacement . That's how
07:52	far he traveled relative to his in his initial position
07:58	. So hopefully this example gave you a good understanding
08:01	between the difference of distance and displacement . So remember
08:05	, distance is a scale of quantity . It's always
08:08	positive and displacement is a vector . It has both
08:13	magnitude and direction . Now let's talk about the difference
08:17	between speed and velocity . Remember speed is a scale
08:23	of quantity but velocity is a vector quantity , speed
08:29	has magnitude only , but velocity has both magnitude and
08:32	direction . Let's use s to describe speed as would
08:39	represent the instantaneous speed . That is how fast an
08:42	object is moving at an instant of time . But
08:45	S . Bar represents the average speed . Average speed
08:50	is equal to the distance traveled divided by the time
08:57	that was elapsed , average velocity is equal to displacement
09:06	over time . So therefore speed is always positive .
09:16	Velocity can be positive or negative depending on the direction
09:23	. So average velocity is basically the velocity calculated over
09:27	a time interval . Now let's use an example to
09:31	calculate speed . I mean average speed and average velocity
09:34	. So let's say we have a particle and this
09:37	particle Travels 100 m east And then it turns around
09:46	and travels 150 m west and it does all of
09:53	this in five seconds , calculate the average speed and
10:00	the average velocity of this particle Over this 5 2nd
10:05	time interval . So to calculate the average speed first
10:11	, we need to calculate the total distance traveled .
10:15	This particle traveled a total distance of 250 m .
10:20	It traveled 100 m east and then 100 and 50
10:23	m west . So if you add those two numbers
10:26	, you get a total distance of 250 m .
10:30	Now it did this in five seconds to 50 ,
10:33	divided by five is 50 . So the average speed
10:38	Of this particle is positive 50 m/s . Now the
10:46	average velocity is different because the displacement is different .
10:52	Now there's two ways in which we can calculate the
10:54	displacement . We can add up the individual the displacements
11:00	for the individual segments of the problem . The displacement
11:04	for the first part is a 100 m And the
11:07	displacement for the second part is -150 . If we
11:11	add 100 plus negative 150 we get In that displacement
11:16	of -50 . The other way in which you can
11:20	calculate displacement is by taking the final position and subtracted
11:25	by the initial position and it helps to draw a
11:28	number line for this . So the particles started opposition
11:32	zero . And during the first part of this traveled
11:38	It was at position 100 And then it traveled 150
11:42	m west , so it ended at position -50 .
11:48	So its final position was negative 50 Minour its initial
11:53	position of zero , Which will still be negative 50
11:58	. I mean you could do it that way too
11:59	, but I prefer to simply add the displacement of
12:04	each individual part of the problem to get the final
12:07	displacement . So the final displacement is negative 50 m
12:14	Divided by a time of five seconds . So negative
12:17	50 divided by five . The average velocity Calculated is
12:23	negative 10 m/s . So as you can see the
12:29	average speed and the average velocity is not always the
12:34	same , it's going to be the same if the
12:39	person or the object travels in one direction , but
12:42	if there's any change in direction , like when this
12:45	particle you know decided to go west , that's when
12:50	the average speed and the average velocity will be different
12:55	. So keep in mind the average speed doesn't have
12:58	to equal the average velocity . Sometimes they are equal
13:02	to each other , but it's not always the case
13:07	now for instantaneous speed that is the absolute value of
13:11	instantaneous velocity . The instantaneous velocity tells you the velocity
13:16	at an instant of time where is the average velocity
13:21	basically tells you the average over an interval . So
13:25	remember average velocity is the displacement over time and displacement
13:30	is the final position minus the initial position divided by
13:34	t now instantaneous velocity in order to calculate it .
13:41	You need to use limits . So it's to limit
13:44	as the change in time goes to zero . And
13:48	it's basically the displacement over time or the change of
13:53	position over the change in time . So that's how
13:57	you would calculate instantaneous velocity . Which we really won't
14:05	go over that in this video . But for those
14:07	of you who want the formula , that's what it
14:09	is . And whenever you see delta this triangle ,
14:15	it represents the change of something in this case the
14:19	change in position . So delta X . Is the
14:22	final position minus the initial position . So basically it's
14:26	the displacement along the X axis . But you can
14:31	also have displacement along the y axis . In this
14:34	case it will be the final position along the y
14:36	axis , modest initial position . So whenever you see
14:41	the symbol D , you could describe it using distance
14:46	or displacement . Now let's talk about some formulas that
14:52	you need to be familiar with when an object is
14:59	moving with constant speed . Typically this is the formula
15:05	that you need to work with . D . Is
15:08	equal to VT . So in this equation D could
15:14	represent distance or displacement depend on how you use the
15:18	problem V . You could use speed or velocity .
15:25	It's going to work if you're dealing with constant speed
15:29	. But understand this story . If you're using V
15:32	. S speed then D is going to be the
15:33	distance . If you're using V with reference to velocity
15:38	D . Is going to represent the displacement . Now
15:42	for objects moving at constant speed , you need to
15:45	know that the instantaneous velocity is the same as the
15:51	average velocity because the velocity is not changing . So
15:55	whether you use V or V bar it will have
15:58	the same effect . Now let's talk about when objects
16:05	are moving with constant acceleration . And by the way
16:15	, remember if you're using displacement , displacement is the
16:22	final position minus the initial position . But if you're
16:25	dealing with distance , you don't have to worry about
16:26	that . So just keep in mind if you're using
16:29	Diaz displacement or distance now for constant acceleration , we
16:38	have some formulas that we need to take into account
16:45	for cause acceleration , D . Is equal to V
16:48	bar times T . And the reason for that is
16:52	that V doesn't equal V bar when this acceleration anytime
17:01	this acceleration , that means the velocity is changing ,
17:05	acceleration tells you how fast the velocity is changing .
17:09	And if the velocity is changing then the instantaneous velocity
17:13	and the average velocity for most of the time won't
17:17	be the same . So that's why we need to
17:26	use V bar . Now the average velocity is the
17:32	average of the initial velocity And the final velocity .
17:37	So it's basically the some of the initial of the
17:39	final velocity divided by two . Or you can fight
17:43	it this way , 1/2 the initial plus of the
17:46	final . So if we were to replace V Bar
17:52	with this expression we would get The displacement is equal
17:56	to 1/2 the initial plus of the final times .
18:02	T . So remember if you're using Diaz distance ,
18:09	then the initial is the initial speed . The final
18:12	is the final speed . But if you're using ds
18:15	displacement , v . Initio is your initial velocity the
18:18	final is your final velocity . So let's rewrite that
18:25	equation here . This is one of those equations that
18:29	you want to write in your list of equations .
18:33	If you have a test coming up now there's some
18:36	other equations that we're going to add to this list
18:40	. So we said that average velocity is the displacement
18:45	, which is final position minus initial position divided by
18:48	T . Well , if you would have rearranged that
18:52	equation , You'll get this one final position is equal
18:58	to initial position plus the average velocity times T .
19:07	You can also get it from this equation . D
19:10	is equal to V . Bar times T . And
19:17	D being the displacement is x final minus x .
19:20	Initial . And so if you were to move this
19:24	to that side , you would end up with the
19:29	same equation . So there's many ways in which you
19:31	can derive that equation . So that's the next one
19:34	that you want to have in your list . Now
19:40	we said that average velocity is the displacement or the
19:45	change of position divided by the time , average acceleration
19:49	is the change in velocity divided by time . It's
19:54	the final velocity minus the initial velocity divided by T
19:59	. So if you re arrange that equation , you
20:01	get something similar to the one that we have above
20:04	V . final is equal to the initial plus 80
20:11	. Now there are some other equations that we can
20:13	add to the list . Another one is the final
20:17	squared is equal to the initial squared Plus two times
20:21	a . D . And then there's this one ,
20:27	displacement is equal to the initial T . Plus one
20:31	half at squared . And any time you see the
20:35	letter D . You can replace that with X .
20:37	Final minus x . Initial . So if we were
20:41	to substitute X final minus X . Initial for D
20:46	. And then move X initial to the right side
20:48	of the equation . We'll get this equation . Final
20:52	position is equal to initial position plus the initial T
21:00	. Plus one half a . T squared now because
21:05	this is X . Where dealing with motion along the
21:08	X axis . But when you go into project that
21:12	motion you can apply motion along the Y axis as
21:17	well . So you might see this variation of this
21:20	formula Y final equals Y initial plus V . Y
21:25	. Initial T . So that's initial velocity but in
21:28	the Y direction plus one half A . T squared
21:32	. Where A . Is like the gravitational acceleration in
21:36	the right direction . So just understand that you can
21:43	apply these formulas in the X . Direction or in
21:46	the Y . Direction . So it might be a
21:49	lot to keep track of . But as you begin
21:51	to work problems , uh the use of these formulas
21:54	will begin to make more sense . But you may
21:57	want to write these down for your reference now ,
22:00	before we work on a few example problems , there's
22:03	something I do want to mention when dealing with constant
22:05	speed , you could use this formula X final is
22:09	equal to X initial plus VT . So you don't
22:13	need to survive V Bar because V and V bar
22:16	are the same when dealing with constant speed . But
22:21	for constant acceleration , it's good to keep in mind
22:26	that this represents average velocity . So it's better to
22:33	write V Bar instead of the so there's no confusion
22:35	because these two , they're not necessarily the same when
22:39	dealing with constant acceleration . So just be aware of
22:43	that little detail , Let's start with this one .
22:47	A bus is traveling at a constant speed of 40
22:50	m/s . How many hours will it take to travel
22:55	? A distance of 200 miles ? So what formula
23:00	do we need to use ? So we're given the
23:04	speed which we can use as a symbol , V
23:08	or S . And we know the distance which is
23:15	200 miles . The only form of that we could
23:21	use since the bus is moving at constant speed is
23:26	this one . D . Is equal to VT .
23:30	Now , before we use that formula , we need
23:33	to take a look at the units . Here we
23:35	have meters per second and here we have mouse and
23:40	we want to find a time in hours . So
23:43	we're looking for tea , the units they don't match
23:48	. So before we could use the formula , we
23:50	need to convert the units into an appropriate form .
23:54	So what do you recommend that we need to do
23:58	? We have the distance in mouse and we want
24:01	to find a time in hours . The best thing
24:04	we can do is convert the speed from meters per
24:07	second , two mph . Once we do that ,
24:13	then it will match with the unit hours and the
24:18	unit mouse . So we can now use that formula
24:21	. So let's go ahead and convert 40 m/s ,
24:25	two mph . So let's get an overview of how
24:31	we're gonna do this . Let's convert meters , two
24:35	km and then kilometers to mouse And then we're gonna
24:39	convert seconds , two minutes and then minutes to hours
24:47	. So in order to convert miles to kilometers ,
24:49	you need to know the relationship between the two .
24:52	One kilometer is equivalent to 1000 m . I meant
24:58	to say m instead of mouse , but that's how
25:00	you can go from m two km . Now ,
25:04	these units cancel . Now let's convert from kilometers to
25:08	mouse . So we need to put the unit kilometers
25:13	on the bottom miles on top so that these units
25:17	will cancel . You need to know that One mile
25:22	is equal to 1.609 km . So now we have
25:27	the unit Mouse , let's convert seconds , two minutes
25:31	since we have seconds on the bottom , we want
25:33	to put seconds on top and then minutes on the
25:36	bottom , one minute is equal to 60 seconds .
25:41	So now we can cross out this unit and now
25:47	let's convert minutes to hours . one hour is equivalent
25:51	to 60 minutes . And so we could cancel the
25:55	unit minutes . So now let's do the math .
26:03	We're going to multiply by all the numbers that are
26:06	on the top . But we're gonna divide by the
26:08	numbers on the bottom , So it's gonna be 40
26:11	, divided by 1000 . Take that result divided by
26:14	1.60.9 And then multiply that by 60 and then buy
26:19	another 60 . So for the velocity or rather than
26:26	speed You should get 89 .5 If you run it
26:34	MPH which we can write mph . So that's how
26:40	we can convert meters per 2nd 2 mph . Now
26:46	let's go ahead and finish this problem . So let's
26:49	use the formula D . Is equal to V .
26:51	T . In this case D . is going to
26:55	represent the distance which is 200 miles . V .
27:00	is going to be the speed which is 89 0.5
27:06	MPH . And then now we can calculate team so
27:12	he gets here by itself . We need to divide
27:14	both sides By 89.5 MPH . 200 divided by 89.5
27:36	gives us this answer to point 23 hours . Now
27:44	. Looking at the units , we can see that
27:46	the unit miles cancel leaving behind the unit hours ,
27:50	which is , well , we want the final answer
27:53	to be in . So when dealing with these problems
27:57	, always check to make sure that the units match
28:01	. If they don't match , you need to convert
28:03	one unit into another . So just keep that in
28:06	mind . Now let's move on the part B of
28:14	this problem . If the bus moved from a position
28:17	that is 50 miles east of city XY . Z
28:22	. to a position that is 90 miles west of
28:25	city X Y . Z . In five hours ,
28:26	what is the average velocity of the bus during that
28:29	time in the vote ? So let's say this is
28:36	City X , Y Z . So the bus was
28:43	initially , let's say opposition , eh , which is
28:48	50 miles east of city . Ex wise , actually
28:54	, that's wes , I take that back , Let's
29:03	say city X , YZ is at the origin position
29:06	zero . So the bus was 50 miles east of
29:10	the city . So it was initially here . So
29:15	this is going to be X initial , that's the
29:17	initial position of the bus and then it moved To
29:23	a position that is 90 miles west of that city
29:28	. So this is gonna be -90 . That's the
29:31	final position of the bus . So the buses traveling
29:38	west , which means that the average velocity should be
29:42	a negative value . Now , how can we calculate
29:48	the average velocity ? What formula do we need to
29:51	use ? It really helps to write down everything .
29:54	We know that X initial is 50 X . Final
29:59	is negative 90 and the units are in mouse and
30:08	then we have the time , which is five hours
30:12	. We want to calculate the average velocity . Well
30:16	we know that average velocity is displacement over time and
30:20	displacement is the change in position . X final minus
30:24	X . Initial whenever you're dealing with motion along the
30:29	X axis and then divided by the time . So
30:33	the final position is -90 . The initial position is
30:37	positive 50 . The time is five hours So -90
30:45	- positive 50 . That is -140 . So the
30:52	change in position or a displacement that is negative 140
30:58	miles . And we're going to divide that by five
31:00	hours . So 1 40 divided by five is 28
31:05	. Thus the average velocity , it's going to be
31:09	negative 28 MPH which we can right as mph .
31:18	So that's how we can calculate the average velocity for
31:21	this particular problem .

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