Distance, Displacement, Average Speed, Average Velocity - Physics - Free Educational videos for Students in K-12

Distance, Displacement, Average Speed, Average Velocity - Physics - By The Organic Chemistry Tutor

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00:0-1	in this video we're going to focus on distance displacement
00:04	, average speed and average velocity . But let's start
00:08	with the number line . So you can see The
00:11	difference between the two . So let's start opposition negative
00:24	three . And if we travel to a final position
00:30	of two , what is the distance traveled ? And
00:34	what is the displacement ? The distance traveled is simply
00:40	five units . We traveled five units to the right
00:46	now . What about the displacement ? The displacement is
00:50	also five . In this case whenever you move to
00:56	the right and if you don't change the direction ,
00:58	the distance and the displacement is the same , I'm
01:01	going to use D . For distance and delta X
01:05	. To represent displacement . Now let's say if we
01:11	start at position for and we're traveling to position negative
01:22	too . What is the distance and what is the
01:26	displacement ? In order to find the displacement ? It's
01:31	basically the change in X . It's the final position
01:34	minus the initial position . The final position is negative
01:39	two . That's where we stopped . That's the end
01:42	position . We started at four . So that's the
01:45	initial position . So it's negative too - the initial
01:50	position of four . So the displacement is negative .
01:54	six displacement is a vector . It can be positive
01:58	or it can be negative direction is important . But
02:02	now what is the distance that we traveled ? The
02:05	distance is always a positive value . We traveled six
02:10	units to the left . So when describing distance ,
02:13	distance is a scale of quantity . It's always positive
02:17	but displacement can be positive or negative . So anytime
02:21	you travel towards the right displacement is positive . If
02:25	you travel towards the left , displacement is negative but
02:28	distance is always positive regardless if you're moving to the
02:32	right , what's the left ? Yeah . Now let's
02:47	say if we started at negative four and we travel
02:53	It's a position five . And then after that We
02:56	traveled from position 5 to position negative too . What
03:02	is the distance traveled by the object or the particle
03:06	? And what is the displacement ? So let's calculate
03:10	the distance . 1st , we traveled from negative 4
03:14	to 5 . That's nine units to the right .
03:17	And then we traveled from 5 to -2 . That's
03:20	seven units to left . To find the distance .
03:24	You simply add those values . So we traveled a
03:26	total of 16 units now to find the displacement .
03:31	There's two ways we can do that . We could
03:35	use the equation . It's the final position minus initial
03:38	position . So the final position is -2 . The
03:42	initial position was -4 . So it's negative too ,
03:47	minus negative four , Which is the same as -2-plus
03:52	4 . Whenever you have to negative signs next to
03:54	each other you can make it a positive sign .
03:57	So therefore it's positive too . So the distance Travel
04:02	to 16 units but the displacement Is only positive two
04:06	units . And it makes sense because we started at
04:11	negative four and then we ended At -2 . So
04:16	the net effect is we only travel to units to
04:19	the right if you just compare the start and the
04:23	finish where you started , where you ended . So
04:26	that's our net , that's the displacement . It's just
04:29	the distance between the final position and the initial position
04:35	. Now let's break it up into two parts .
04:39	The displacement for the first part is positive nine .
04:42	Because we traveled nine units to the right . The
04:45	displacement for the second part is -7 because we traveled
04:49	seven years to left . If you add those two
04:52	numbers positive nine And -7 , you get the net
04:56	displacement of positive too . Which is what we have
05:00	Now . If you want to calculate the distance ,
05:02	you simply make these two numbers positive Instead of using
05:06	negative seven use positive seven . So nine plus seven
05:08	will give you this value 16 . So hopefully this
05:11	helps you to distinguish distance from displacement . So let's
05:15	say if a car travels 10 miles east and then
05:22	six miles west , calculate the distance and the displacement
05:28	of the vehicle . So if you want to calculate
05:31	the distance , simply add the two numbers and make
05:35	it positive . So it's just gonna be 10 plus
05:37	six . So the total distance traveled is 16 miles
05:41	. Now if you want to find the displacement ,
05:44	you need to do it this way . When the
05:47	car traveled 10 miles to the right the displacement is
05:51	positive . And then when it travels six months to
05:53	the left it's negative . So you add these values
05:56	10 plus negative six . So then that displacement Is
06:00	four miles to the right . So the vehicle it
06:05	started at this point , but then it ended at
06:09	this point . So the net result is that it
06:11	traveled four mi to the right . And so that's
06:15	the net displacement of the vehicle . Now let's say
06:21	if a person walks three miles east And then four
06:28	miles north , calculate the distance and the net displacement
06:34	traveled by the person . So to find a total
06:37	distance , you simply need to just add everything .
06:40	He traveled three miles east and four miles north .
06:43	So he traveled a total distance of three plus four
06:46	or seven miles . Now to find the displacement ,
06:51	we need to find the distance between where he started
06:55	and where he ended . So this is where he
06:59	started and this is his finished or where he ended
07:03	. So we got to find the direct distance between
07:06	those two points . And that's going to give us
07:08	the displacement . So let's call it delta X .
07:13	The displacement . You can call it something else if
07:17	you want , you can call it S . Or
07:19	delta Y . But you know , since we're not
07:22	traveling any X . Direction , were also traveling in
07:24	the Y direction , let's just call it S .
07:31	We need to do is use the pythagorean theorem .
07:34	So in this case we know that C square ,
07:37	the hypothesis is equal to a squared plus B squared
07:40	, So S squared , it's going to be three
07:43	squared plus four square . So that's gonna be nine
07:46	plus 16 , Which adds up to 25 . And
07:52	so the displacement , It's gonna be the square root
07:55	of 25 or five . So the net effect is
08:00	that the person traveled five mouse in that direction ,
08:05	Even though he actually walked a distance of seven mouse
08:11	. So keep in mind the displacement is the distance
08:13	between your final position and the initial position . Now
08:19	let's work on some problems . An object moves from
08:22	position negative eight To 12 . And then to position
08:27	-20 . What is the total distance traveled by the
08:30	object ? And what is the displacement ? So for
08:33	these problems , feel free to pause the video and
08:35	work it out yourself and then composite to see if
08:38	you have the right answer . So let's go ahead
08:40	and begin . So we're going to start at negative
08:49	eight . And then The object travels from a position
08:54	of negative 8 - 12 . So how far did
09:00	the object traveled at this time ? Let's find the
09:06	displacement just for the first part of the trip .
09:10	To find the displacement . It's the final position minus
09:14	the initial position . So the final position is 12
09:17	. Just for the first part minus an initial position
09:20	of -8 . 12 minus negative eight Is the same
09:23	as 12-plus 8 . So right now the object traveled
09:28	A distance of 20 m to the right now .
09:33	What about the second part of the trip ? How
09:41	far did the object traveled at that point ? So
09:47	the object starts at position 12 and then moves towards
09:51	position -20 , which should be somewhere over there .
09:58	So let's calculate the displacement just for the second part
10:01	of the trip . So using the same formula DELTA
10:05	X . Is X final minus X . Initial .
10:10	So for the second part , the final position is
10:13	negative 20 - the initial position of positive 12 .
10:18	So that's negative 20-12 , Which is -32 . The
10:24	negative value means that the object is moving towards the
10:28	left . So now how can we use this information
10:35	to find a total distance and then that displacement for
10:39	the entire trip ? So to find a total distance
10:42	, simply make everything positive . He traveled 20 m
10:47	to the right And then 32 m to the left
10:51	. So when dealing with distance , everything is positive
10:54	. So he traveled or the object rather traveled ,
10:57	a total distance of 52 m is 20 Plus 32
11:01	. Distance is always positive . Now , if you
11:04	want to find the nut displacement for the entire trip
11:08	, During the first part , it's positive 20 And
11:12	during the second part it's -32 . So you got
11:15	to incorporate the negative sign , So it's 20-32 ,
11:19	Which is -12 . So the displacement is negative 12
11:23	, but the total distance traveled is 52 . Now
11:28	you can also find the displacement using this formula .
11:33	The initial position for the entire trip starts at -8
11:38	. The final position for the entire trip ends negative
11:44	20 . So this can give us the net displacement
11:48	for everything rather than just break it up into two
11:50	parts . So the final position is negative 20 minus
11:55	the initial position of negative eight . So this is
11:57	negative 20 plus eight , Which turns out to be
12:01	-12 . So there's many different ways that you can
12:05	employ to calculate the net displacement of the object number
12:10	two . Sally travels 50 m west And then 120
12:15	m south . How far did sally travel ? And
12:19	what is her net displacement ? So go ahead and
12:22	try this problem . So let's try a picture .
12:25	So let's say she started here . She traveled 50
12:29	m west and then 120 m south . So if
12:37	we call this is east , this is west ,
12:40	this is north and this is self , so this
12:50	is where she started and this is her final position
12:54	. So therefore , if we draw a line from
12:56	start to finish , the left of that line will
13:00	represent her net displacement , which we can call Delta
13:07	S . But let's find the total distance first .
13:11	So she traveled 50 m west and then 100 20
13:15	m south . So it's 50 plus 1 20 .
13:19	So a total distance Is 170 m . That's how
13:23	far she actually traveled . Now to find the displacement
13:29	. We just need to find the left of the
13:30	hypothesis of the triangle . So Delta asked squared ,
13:35	it's going to be a squared plus B squared or
13:37	50 squared Plus 120 sq . So 50 square ,
13:44	that's 50 times 50 , that's 2500 And 120 times
13:54	120 Is 14,400 . So if we are these two
14:00	numbers , this is going to give us 169,000 ,
14:06	I mean 16,900 . So now let's take the square
14:11	root of both sides . The Square Root of 16,900
14:20	is 130 . So that is sally's net displacement .
14:28	That's the distance between her initial position and her final
14:32	position . And so that's the answer for part B
14:38	. Let's work on # three . Megan walks 100
14:42	m east and then travel 70 m north , followed
14:45	by 140 m east , calculate the total distance and
14:50	that displacement . So she travels a 100 m east
14:58	And then she's gonna travel 70 m north And then
15:03	140 m east . So to find the total distance
15:09	, We just need to add the numbers 100 plus
15:12	70 plus 1 40 70 plus 1 40 . That's
15:18	2 10 plus 100 that street . And so that's
15:23	a total distance traveled by Megan . Now to find
15:27	the net displacement . Let's draw a line from her
15:30	initial position to her final position . How can we
15:37	find that distance ? So what I'm going to do
15:41	is I'm going to redraw it . I'm going to
15:44	draw this vector first and this one right after it
15:49	. So this problem is equivalent to this problem ,
15:52	let's say if she traveled 100 m east And then
15:55	another 1 40 m east And then 70 m north
16:01	. Now notice that we have a right triangle .
16:04	Our goal is to find the left of this green
16:07	line , which we can call Delta S . So
16:14	if we add these two numbers , 100 plus 1
16:16	40 that's 2 40 Going east and then 70 going
16:23	north . So now we could use the protagonist here
16:25	. Um So Delta S is going to be the
16:28	square root of 70 squared Plus 240 square . And
16:35	you can type it in exactly the way you see
16:36	it in your calculator . So you should get 250
16:48	. That is her net displacement . That's the distance
16:51	from the initial position to her final position . It's
16:56	250 m in that direction . Here's another example Jared
17:03	Walks 120 m east And then 150 m south And
17:16	then 40 m west . Let's start by finding the
17:22	total distance . So all we need to do is
17:26	add 120 plus 1 50 plus 40 . 1 20
17:31	plus 1 50 that's 2 70 plus 40 . So
17:35	that's 310 m traveled . So it's very simple to
17:38	find a total distance . You just got to add
17:40	the numbers . Now we got to find the displacement
17:43	which is the distance between the initial and the final
17:46	position . So how long is the blue line ?
17:52	So here's what we're gonna do . We need to
17:54	turn this into a triangle . So if he traveled
17:59	120 m east and then 40 m west in the
18:03	X . direction . His net displacement is only 80
18:07	m to the right . So this is positive on
18:15	20 and this is negative 40 in the X .
18:17	direction . If you add those two , that's going
18:21	to give you positive , aiding any extra action in
18:24	the Y direction , there's only one number , 1
18:27	50 . So we can't do anything with that .
18:31	So now we can find the high partners . So
18:36	the displacement is going to be the square root of
18:39	a squared plus B squared or 80 square Plus 150
18:44	Squared . So if you type that in exactly what
18:53	you see it . This should give you 1 70
18:56	. It's based on the 8 1517 Right Triangle .
19:00	So that's the net displacement of jerry . It's helpful
19:06	to know the special right triangles because if you know
19:09	them , you can quickly solve certain pythagorean theorem related
19:13	problems . The first one Is the 345 right triangle
19:19	. So we see another triangle where this is 30
19:22	and this is 40 . You know , the Missus
19:24	side has to be 50 . So there's a 345
19:27	right triangle , There's the 5 12 13 Triangle .
19:32	So this is 50 and this is 120 . The
19:35	missing side is 1:30 . The next one is the
19:38	8 15 17 right triangle . And then the 7:24
19:44	25 by travel . Those are the most common ones
19:46	that you may encounter . Some other rare ones you
19:49	might see Is the 9 40 41 Triangle and the
19:53	1160 61 . Other than that , I doubt you
19:59	see anything else . But you can see any ratio
20:03	of these numbers . For example The 345 triangle .
20:06	If you multiply by two You'll get to 6810 triangle
20:10	. So this is 60 and this is 80 .
20:13	The Missus side is 100 . So if you know
20:15	these numbers then you could solve a lot of common
20:20	right triangle problems quickly . Now let's talk about the
20:24	difference between average speed and average velocity as well as
20:28	how to calculate them . Now keep in mind ,
20:31	speed is a scale of quantity . It has magnitude
20:35	only and no direction velocity is a vector quantity .
20:40	It has both magnitude and direction . So velocity is
20:43	basically speed with direction . Now if you want to
20:47	calculate average speed when the k average with a bar
20:51	on top of the s average speed is equal to
20:54	the total distance traveled divided by the total time .
21:01	Average velocity is equal to the displacement divided by the
21:08	time . So displacement is the vector . Therefore velocity
21:14	will be a vector distance is a scalar quantity .
21:17	So speed will also be scale as well . Now
21:23	speed mhm . That is instantaneous , speed is the
21:28	absolute value of velocity . So what this means is
21:30	that speed will always be positive . Velocity can be
21:34	positive or negative depending on the direction . So for
21:38	instance , let's say if Karen is traveling at a
21:42	speed of 30 mph east , care and speed is
21:49	30 her velocity is 30 mph east . The units
21:59	are the same . So as you can see velocity
22:01	of speed with direction . Now let's say Jim is
22:04	traveling At a speed of 40 mph west . So
22:12	his speed At that instant is positive 40 mph .
22:17	So in both cases , if a person is traveling
22:20	east or west the speed is positive now because this
22:25	person is going in the negative X direction , the
22:28	velocity will not be positive . The velocity it's negative
22:33	. So his velocity is negative 40 mph in the
22:36	direction west . So as you can see , speed
22:42	is always positive , but velocity can be positive or
22:44	negative depending on where a person is going . So
22:48	make sure you understand that concept . Now let's work
22:51	on some practice problems . A car travels a distance
22:55	of 300 miles in six hours . What is the
22:57	average speed of the car ? Mm So we could
23:02	use this form of the average speed . It's going
23:04	to be the distance traveled divided by the time .
23:10	Mhm . So the distance is 300 miles , The
23:17	time is six hours . So what's 300 divided by
23:22	6 ? Well we know that 30 divided by six
23:26	is 5 . So 300 divided by six is 50
23:30	. So we get 50 MPH . Or we can
23:33	fight as 50 mph . Yeah , so that's it
23:37	for the number one . That's how you can calculate
23:40	the average speed . And you can get the units
23:43	by looking at the units of distance and time .
23:47	Let's go ahead and solve these two problems . A
23:50	car travels at an average speed of 40 ft/s .
23:54	How many miles will it travel in five hours ?
24:00	So we could use the formula D . Is equal
24:02	to r . T . Distance equals rate times time
24:06	or D . Equals VT . Where V . Is
24:09	the rate average speed , average velocity . So it's
24:13	the it will lead to the same answer regardless of
24:16	which of those two formulas you decide to use .
24:20	But let's write down what we know . So we
24:25	know the average speed is 40 ft/s . We know
24:31	the time is five hours . Our goal is to
24:35	calculate the distance now . We don't want to plug
24:38	it in into that formula right now because the units
24:42	they don't match . So what should we do here
24:49	? We want the distance in mouse and we have
24:52	the time in hours . So in order for the
24:55	units to match , we need to convert the average
24:57	speed from feet per 2nd , 2 mph . And
25:01	then we could plug everything into the formula . So
25:04	let's go ahead and do that . So we have
25:07	40 ft/s . The first thing I need to be
25:11	familiar with is the conversion between miles and feet .
25:14	One mile is 5280 ft . So I'm going to
25:19	put the unit feet on the bottom and one mile
25:23	on top so that the unit feet will cancel .
25:27	All right Now let's convert seconds in two minutes .
25:32	There's 60 seconds in one minute . And so we
25:37	could cross out the unit seconds . Mhm . And
25:44	there is 60 minutes in a single hour . So
25:51	that's how we can convert from feet per second two
25:54	mph . Now let's go ahead and plug this in
25:57	. So it's 40 divided by 50 to 80 Times
26:00	60 times 60 . So the new value of the
26:08	is 27 .27 repeating MPH . Yeah . So now
26:26	let's go ahead and calculate the distance . The distance
26:28	is going to be , The speed multiplied by the
26:30	time . So v . 27 0.27 We have miles
26:40	on top hours on the bottom and then we're gonna
26:43	multiply by T . Which is five hours . So
26:49	we can see that the unit hours will cancel .
26:54	So 27.27 repeat in times five . That's 136.36 repeating
27:03	mouse . So that's how many miles this car will
27:11	travel in five hours . Now let's move on to
27:16	part B . A train is moving at 45 km/h
27:24	. How long will it take for the train to
27:26	travel a distance of 20 miles . So let's write
27:30	down what we know . We know the distance is
27:33	20 miles , how long ? So we're looking for
27:36	the time And were given the speed , it's 45
27:43	kilometers per hour . Now we can change the speed
27:51	from km/h , mph . Or we could change the
27:54	distance from miles to kilometers . We can do it
27:59	both ways . But the key is that these two
28:01	, they need to match . So this time let's
28:03	change the distance . The conversion between miles and km
28:13	is this one mouth is 1.609 km . You can
28:19	look this up online as well , So it's 20
28:22	times 1.609 and that gives us A distance of 32.18
28:30	kilometers . So now let's go ahead and use this
28:35	formula D is equal to VT . We need to
28:38	isolate t to do that . Let's divide both sides
28:42	by V . Mhm . So time is distance over
28:52	average speed . So this train is going to travel
28:56	a distance of 32.18 km And its average speed is
29:04	45 km/h . So we could see the unit kilometers
29:09	will cancel and the time is gonna come out in
29:11	hours . So 32 18 Divided by 45 , this
29:21	is point 715 , 1 repeated hours . Now ,
29:27	if we multiply this by 60 minutes per hour ,
29:30	we can get the answer in minutes , which we
29:33	could say it's approximately 42.9 minutes . So that's how
29:41	we could find the time . If we know the
29:43	distance and the average speed

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