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Distance vs Displacement, Speed vs Velocity · Acedra
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Distance vs Displacement, Speed vs Velocity
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0:00
in this video we're going to focus on distance displacement average speed and average velocity but let's start with a number line so you can see the difference between the two
0:20
so let's start at position negative three and if we travel to a final position of two what is the distance traveled and what is the displacement
0:38
the distance traveled is simply 5 units we traveled five units to the right now what about the displacement the displacement is also five
0:55
in this case whenever you move to the right and if you don't change the direction the distance and the displacement is the same i'm going to use d for distance and delta x to represent displacement now let's say if we start at
1:15
position four and we're traveling to position negative two what is the distance and what is the displacement in order to find the displacement it's basically the change in x it's the final position minus the initial position the final position is negative two that's where we stop that's the end position we started at four so that's the initial position so it's negative two minus the initial position of four so the displacement is negative six displacement is a vector it can be
1:58
positive or it can be negative direction is important but now what is the distance that we traveled the distance is always a positive value we traveled six units to the left so when describing distance distance is a scalar quantity it's always positive but displacement can be positive or negative so anytime you travel towards the right displacement is positive if you travel towards the left displacement is negative but distance is always positive regardless if you're moving to the right
2:33
or to the left
2:47
now let's say if we started at negative four and we travel to position five and then after that we travel from position five to position negative two what is the distance traveled by the object or the particle and what is the displacement so let's calculate the distance first we traveled from negative four to five that's nine units to the right and then we travel from five to negative two that's seven units to the left to find the distance you simply add those values so we traveled a total of
3:28
sixteen units now to find the displacement there's two ways we can do that we could use the equation it's the final position minus the initial position so the final position is negative two the initial position was negative four so it's negative two minus negative four which is the same as negative two plus four whenever you have two negative signs next to each other you can make it a positive sign so therefore it's positive 2. so the distance traveled is 16 units but the
4:04
displacement is only positive 2 units and it makes sense because we started at negative 4 and then we ended at negative two so the net effect is we only travel two units to the right if you just compare the start and the the finish where you started where you ended so that's our net that's the displacement it's just the distance between the final position and the initial position now let's break it up into two parts the displacement for the first part is positive nine
4:43
because we traveled nine units to the right the displacement for the second part is negative seven because we traveled seven units to left if you add those two numbers positive nine and negative 7 you get the net displacement of positive 2 which is what we have now if you want to calculate the distance you simply make these two numbers positive instead of using negative 7 you use positive 7 so 9 plus 7 will give you this value 16. so hopefully this helps you to distinguish
5:13
distance from displacement so let's say if a car travels 10 miles east and then six miles west calculate the distance and the displacement of the vehicle so if you want to calculate the distance simply add the two numbers and make it positive so it's just going to be 10 plus 6 so the total distance traveled is 16 miles now if you want to find the displacement you need to do it this way when the car travels 10 miles to the right the displacement is positive and then
5:52
when it travels six months to the left it's negative so you add these values 10 plus negative six so then that displacement is four miles to the right
6:04
so the vehicle it started at this point but then it ended at this point so the net result is that it traveled four miles to the right and so that's the the net displacement of the vehicle now let's say if a person walks three miles east and then four miles north calculate the distance and the net displacement traveled by the person so to find the total distance you simply need to just add everything he traveled three miles east and four miles north so he traveled a total distance of three
6:46
plus four or seven miles now to find the displacement we need to find the distance between where he started and where he ended so this is where he started and this is his finish point or where he ended so we got to find the direct distance between those two points and that's going to give us the displacement so let's call it delta x the displacement you can call it something else if you want you can call it s or delta y but you know what since we're not traveling in the x direction we're also
7:24
traveling in the y direction let's just call it s what we need to do is use the pythagorean theorem so in this case we know that c squared the hypotenuse is equal to a squared plus b squared so s squared is going to be three squared plus four squared so that's going to be nine plus 16 which adds up to 25 and so the displacement is going to be the square root of 25 or five so the net effect is that the person traveled five miles in that direction even though he actually walked a distance
8:08
of seven miles so keep in mind the displacement is the distance between your final position and the initial position now let's work on some problems an object moves from position negative 8 to 12 and then to position negative 20. what is the total distance traveled by the object and what is the displacement so for these problems feel free to pause the video and work it out yourself and then unpause it to see if you have the right answer so let's go ahead and begin
8:46
so we're going to start at negative 8 and then the object travels from a position of negative 8 to 12.
9:00
so how far did the object traveled at this time let's find the displacement just for the first part of the trip to find the displacement it's the final position minus the initial position so the final position is 12 just for the first part minus an initial position of negative eight twelve minus negative eight is the same as twelve plus eight so right now the object traveled a distance of twenty meters to the right now what about the second part of the trip
9:41
how far the object traveled at that point
9:47
so the object starts at position 12 and then moves towards position negative 20 which should be somewhere over there so let's calculate the displacement just for the second part of the trip so using the same formula delta x is x final minus x initial so for the second part the final position is negative 20 minus the initial position of positive twelve so that's negative twenty minus twelve which is negative thirty two the negative value means that the object is moving towards the left
10:33
so now how can we use this information to find the total distance and the net displacement for the entire trip so to find the total distance simply make everything positive he traveled 20 meters to the right and then 32 meters to the left so when dealing with distance everything is positive so he traveled or the object rather traveled a total distance of 52 meters is 20 plus 32 distance is always positive now if you want to find the net displacement for the entire trip
11:08
during the first part it's positive 20. and during the second part it's negative 32 so you got to incorporate the negative sign so it's 20 minus 32 which is negative 12. so the displacement is negative 12 but the total distance traveled is 52. now you can also find the displacement using this formula the initial position for the entire trip starts at negative eight the final position for the entire trip ends at negative 20. so this can give us the net displacement
11:48
for everything rather than just breaking up into two parts so the final position is negative 20 minus the initial position of negative eight so this is negative twenty plus eight which turns out to be negative twelve so there's many different ways that you can employ to calculate the net displacement of the object number two sally travels 50 meters west and then 120 meters south how far did sally travel and what is her net displacement so go ahead and try this problem so let's draw a picture
12:26
so let's say she started here she traveled 50 meters west and then 120 meters south if you recall this is east this is west this is north and this is south
12:50
so this is where she started and this is her final position so therefore if we draw a line from start to finish the length of that line will represent her net displacement which we can call delta s but let's find the total distance first so she traveled 50 meters west and then 120 meters south so it's 50 plus 120 so her total distance is 170 meters that's how far she actually traveled now to find the displacement we just need to find the length of the hypotenuse of the triangle
13:34
so delta s squared is going to be a squared plus b squared or 50 squared plus 120 squared so 50 squared that's 50 times 50 that's twenty five hundred
13:52
and a hundred twenty times 120 is fourteen thousand four hundred so if we add these two numbers this is gonna give us a hundred and sixty nine thousand i mean sixteen thousand nine hundred so now let's take the square root of both sides
14:18
the square root of sixteen thousand nine hundred is a hundred and thirty so that is sally's net displacement that's the distance between her initial position and her final position and so that's the answer for part b let's work on number three megan walks 100 meters east and then travels 70 meters north followed by 140 meters east calculate the total distance and net displacement so c travels 100 meters east and then she's going to travel 70 meters north and then 140 meters east
15:07
so to find the total distance we just need to add the numbers 100 plus 70 plus 140 70 plus 140 that's 210 plus 100 that's 310. so that's the total distance traveled by megan now to find the net displacement let's draw a line from her initial position to her final position
15:37
how can we find that distance so what i'm going to do is i'm going to redraw it i'm going to draw this vector first and this one right after it so this problem is equivalent to this problem let's say if she traveled 100 meters east and then another 140 meters east and then 70 meters north now notice that we have a right triangle our goal is to find the length of this green line which we can call delta s
16:14
so if we add these two numbers 100 plus 140 that's 240 going east and then 70 going north so now we could use the pythagorean theorem so delta s is going to be the square root of 70 squared plus 240 squared and you could type it in exactly the way you see it in your calculator
16:45
so you should get 250 that is her net displacement that's the distance from the initial position to her final position it's 250 meters in that direction here's another example jared walks 120 meters east
17:11
and then 150 meters south
17:17
and then 40 meters west let's start by finding the total distance so all we need to do is add 120 plus 150 plus 40. 120 plus 150 that's 270 plus 40 so that's 310 meters traveled so it's very simple to find a total distance you just got to add the numbers now we got to find the displacement which is the distance between the initial and the final position so how long is the blue line so here's what we're going to do we need to turn this into a triangle so if he traveled 120 meters east
18:01
and then 40 meters west in the x direction his net displacement is only 80 meters to the right
18:14
so this is positive 120 and this is negative 40 in the x direction if you add those two that's going to give you positive 80 in the x direction in the y direction there's only one number 150 so we can't do anything with that so now we could find the hypotenuse
18:36
so the displacement is going to be the square root of a squared plus b squared or 80 squared plus 150 squared
18:52
so if you type that in exactly the way you see it this should give you 170. it's based on the 8 15 17 right triangle so that's the net displacement of jared it's helpful to know the special right triangles because if you know them you can quickly solve certain pythagorean theorem related problems the first one is the 345 right triangle so if you see another triangle where this is 30 and this is 40 you know the missing side has to be 50. so there's a 345 right triangle
19:29
there's the 5 12 13 triangle so this is 50 and this is 120 the missing side is the next one is the 8 15 17 right triangle and then the 7 24 25 by triangle those are the most common ones that you may encounter some other rare ones you might see is the 9 40 41 triangle and the 11 60 61. other than that i doubt you'll see anything else but you can see any ratio of these numbers for example the three four five triangle if you multiply by two you'll get the six eight ten triangle
20:11
so this is sixty and this is eighty the missing side is a hundred so if you know these numbers then you could solve a lot of common right triangle problems quickly now let's talk about the difference between average speed and average velocity as well as how to calculate them now keep in mind speed is a scalar quantity it has magnitude only and no direction velocity is a vector quantity it has both magnitude and direction so velocity is basically speed with direction
20:47
now if you want to calculate average speed we'll indicate average with a bar on top of the s average speed is equal to the total distance traveled divided by the total time average velocity is equal to the displacement divided by the time
21:12
so displacement is a vector therefore velocity will be a vector distance is a scalar quantity so speed will also be scalar as well now speed that is instantaneous speed is the absolute value of velocity so what this means is that speed will always be positive velocity can be positive or negative dependent on the direction so for instance let's say if karen is traveling at a speed of 30 miles per hour east karen's speed is 30. her velocity is 30 miles per hour east
21:59
the units are the same so as you can see velocity is speed with direction now let's say jim is traveling at a speed of 40 miles per hour west so his speed at that instant is positive 40 miles per hour so in both cases if a person is traveling east or west the speed is positive now because this person is going in the negative x direction the velocity will not be positive the velocity is negative so his velocity is negative 40 miles per hour in the direction west so as you can see speed is always
22:43
positive but velocity can be positive or negative depending on where a person is going so make sure you understand that concept now let's work on some practice problems a car travels a distance of 300 miles in six hours what is the average speed of the car so we could use this formula the average speed is going to be the distance traveled divided by the time
23:14
so the distance is 300 miles the time is 6 hours what's 300 divided by 6 well we know that 30 divided by 6 is 5 so 300 divided by 6 is 50. so we get 50 miles per hour or we can write as 50 mph so that's it for number one that's how you can calculate the average speed and you can get the units by looking at the units of distance in time let's go ahead and solve these two problems a car travels at an average speed of 40 feet per second how many miles will it travel in five hours
24:00
so we could use the formula d is equal to rt distance equals rate times time or d equals vt where v is the rate average speed average velocity so it's the it will lead to the same answer regardless of which of those two formulas you decide to use but let's write down what we know
24:25
so we know the average speed is 40 feet per second we know the time is 5 hours our goal is to calculate the distance now we don't want to plug it in into that formula right now because the units they don't match
24:47
so what should we do here we want the distance in miles and we have the time in hours so in order for the units to match we need to convert the average speed from feet per second to miles per hour and then we could plug everything into the formula so let's go ahead and do that so we have 40 feet per second the first thing you need to be familiar with is the conversion between miles and feet one mile is 5280 feet so i'm going to put the unit feet on the bottom and one mile on top
25:25
so that the unit feet will cancel now let's convert seconds into minutes there's 60 seconds in one minute and so we could cross out the unit seconds
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and there's 60 minutes in a single hour so that's how we can convert from feet per second to miles per hour now let's go ahead and plug this in so it's 40 divided by 5280 times 60 times 60.
26:06
so the new value of v is 27.27 repeating
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miles per hour
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so now let's go ahead and calculate the distance the distance is going to be the speed multiplied by the time
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so v is 27 point we have miles on top hours on the bottom and then we're going to multiply by t which is 5 hours so we can see that the unit hours will cancel so 27.27 repeating times five that's 136.36 repeating miles
27:09
so that's how many miles this car will travel in five hours now let's move on to part b a train is moving at 45 kilometers per hour how long will it take for the train to travel a distance of 20 miles so let's write down what we know we know the distance is 20 miles how long so we're looking for the time and we're given the speed it's 45 kilometers per hour
27:49
now we can change the speed from kilometers per hour to miles per hour or we could change the distance from miles to kilometers we can do it both ways but the key is that these two they need to match so this time let's change the distance
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the conversion between miles and kilometers is this one mile is 1.609 kilometers you can look this up online as well so it's 20 times 1.609 and that gives us a distance of 32.8 kilometers
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so now let's go ahead and use this formula d is equal to vt we need to isolate t to do that let's divide both sides by v so time is the distance over average speed so this train is going to travel a distance of 32.18 kilometers
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and its average speed is 45 kilometers per hour so we can see the unit kilometers will cancel and the time is going to come out in hours
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so 32.18 divided by 45. this is 0.7151 repeating hours now if we multiply this by 60 minutes per hour we can get the answer in minutes which we could say it's approximately 42.9 minutes so that's how we could find the time if we know the distance and the average speed
30:11
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