hey everyone in this video we're going to talk about some Concepts and tips for solving 2D motion and Vector problems and then we'll walk through and solve some different types of problems pause the video at the beginning of each problem and try solving it first when you're done or if you can't figure it out watch the video to see how to work through the problem and check your answer so so what are some of the concepts and skills that we need for 2D motion and Vector problems first we need to know some geometry we use triangles a lot especially right triangles so we need to know how to find the side lengths and angles using the Pythagorean theorem and the right triangle trig functions s cosine and tangent we should also be familiar with finding different angles in sets of triangles or parallel lines and it's not as common but circular motion is two-dimensional motion so we might need to know some geometry about circles next we need to be familiar with coordinate systems we usually use a cartisian coordinate system and we label the axes as X and Y each with a positive and negative direction for some problems we need to work with axes that are rotated or even flipped we also need to know how to use coordinates to describe the position of a point or an object within a coordinate system next we definitely need to know how to work with vectors we should be able to at least roughly draw a vector when given its magnitude and angle we should also be familiar with the different ways that a vector's angle can be described relative to other axes or directions we need to know how to find the components of a vector and how to find its magnitude and angle and we also need to know how to add vectors either graphically by using the tip to tail method or mathematically by adding their components finally we need to be familiar with the one-dimensional kinematic equations we might be working with an object in two-dimensional motion but we're going to be answering questions about its motion either in the direction it's moving if it's a straight line or we're going to find the X and Y components and use the 1D kinematic equations for each Direction those are some of the key Concepts and skills that we need and you can go back and review anything you're not familiar with now let's cover some tips for solving these types of problems tip number one always draw a picture you should always draw a picture at the start of a physics problem but it's especially helpful and sometimes necessary for 2D motion and vectors draw things big enough to fit everything that you want to include use a picture as a helpful reference but remember that it may not be accurate unless you use a ruler and a protractor your vectors and angles probably won't be accurate for example if we're given the magnitudes and angles of vectors A and B and asked to find their resultant vectors C the vectors we draw might not be accurate enough to show us if C points in the positive or negative X Direction so it's always good to draw things out but trust the math instead of your picture draw the X and Y AES if it's useful for the problem but it's not always necessary keeping track of your directions and signs is important but if the problem is simple enough you don't always have to apply the X and Y axes to the situation and remember that when we draw vectors they can be moved around so even if a vector is attached to an object you can draw the vector at the origin to see its angle relative to the X and Y AIS tip number two know the different ways to describe angles pay attention to the wording around angles and make sure you understand exactly which angle is being given or which angle you're being asked to find use the conventional angle where positive angles are count clockwise from the positive xais in these cases if you just working with a vector and it's not related to a physical scenario especially if there's no units if the angle is relative to the positive xais or if the angle you're given is greater than 90° but in most other cases when in doubt just use a positive reference angle that usually just means a positive angle inside a vector's right triangle relative to the xais or the horizontal also make sure your calculator is set to degrees instead of radians if that's what you're working with and tip number three turn a 2d problem into a 1D problem if you're given or you're starting with a 2d vector and you're asked about the motion in the x or y direction or the horizontal or vertical Direction then you can find the components of the vector and use the 1D kinematic equations for each direction or you can start with two 1D motion components and use them to find a 2d Vector remember there's no 2D kinematic equations there's only a combination of 1D kinematic equations all right now let's solve some problems Vector a has has a magnitude of 15 and an angle of 122° vector B has a magnitude of 23 and an angle of 46° what is the magnitude and angle of the resultant Vector of a plus b let's read through this again and find the important information we're given two vectors Vector a has a magnitude of 15 but we're not given any units it has an angle of 122° so let's write down a = 15 and Theta a is 122° Vector B has a magnitude of 23 and an angle of 46° and we'll write that down the problem is asking us to add vectors A and B and find the magnitude and angle of the resultant Vector let's call the resultant Vector C so it's easier to keep track of and write down that we're trying to find C and its angle Theta C next let's draw these vectors to get an idea of what we're trying to find the problem didn't give us any length units and we're told Vector a has an angle greater than 90° that should tell us that these angles are probably measured counterclockwise from the positive xais which is the conventional way to describe angles using a single value so let's draw vectors A and B starting at the origin of the X and Y axis Vector a has an angle of 122° from the positive X AIS and we'll label the magnitude as 15 Vector B has an angle of 46° and a magnitude of 23 remember unless you use some tools the angle and lengths of these vectors won't be accurate so these pictures just help us see the vectors and which directions they're pointing now let's use the tip to tail method to draw their resultant Vector C first we draw a and then we start B at the end of a then we draw Vector C from the start of a to the end of B if we want now we can draw Vector C at the origin to get an idea of its direction however if we didn't draw vectors A and B accurately then C won't be accurate either so we probably don't want to rely on this picture to determine its direction and you can skip this if you want but seeing vectors A and B added together visually is probably helpful okay so what are the steps for adding two vectors and what equations could we use to solve this we learned that if we're adding vectors we can find their individual components components and then add those together to find the components of the resultant Vector then we can use those to find its magnitude and angle so if Vector C equals Vector a plus Vector B then the component CX = ax + BX and c y = a y + b y we know vectors A and B we can find their components then find CX and c y and then use those to find Vector C we do need some other equations to work with vectors like the trig functions but these here will remind us of how to solve the problem so first let's find the components of vector a let's start by using the 122° angle that we're given we learned that the X component ax is equal to the magnitude a Time the cosine of theta we plug in 15 for a and 122° for Theta and we get - 7.95 notice that the X component is negative because it points in the negative X Direction and plugging in this conventional angle takes care of the sign for us next a y = a * the S of theta we plug in the same values for a and Theta and we get 12 72 let's save these components at the top another way we could have found the components is by using the reference angle 180° minus 122° is 58° which is the angle between the vector and the xaxis and if we draw out the components it's the angle inside the triangle between the vector and the X component if we plug in 58 ° then a y is still positive 1272 but now ax is positive 7.95 that's just the length of the bottom side of this triangle but like we said ax points in the negative X Direction and since we're adding vectors we need to include that negative sign for things to work out now let's find the components of vector B just like with Vector a BX = B * the cosine of theta and b y = b * the S of theta now we plug in 23 for B and 46° for Theta and we find that BX is 15.98% components to find the components of vector C CX = ax + BX which is- 7.95 + 15.98% which gives us 29.267562 and we get 30.3 for the magnitude of vector C to find the angle we use the inverse tangent function where Theta is equal to the inverse tangent of the Y component over the X component when we plug in the values of Cy and CX we get 74.7 de for the angle Theta the angle we found is the reference angle between vectors C and its X component which is how the tangent function Works Cy is the opposite side and CX is the adjacent side to this angle but if the problem is asking us for a single value for the angle of vector C we probably want to find its angle counterclockwise from the positive xaxis so is that this angle or a different one if we drew the vectors accurately we would see the vector points in the positive X and positive y directions but we don't need the picture to tell us that because CX and Cy are both positive so if this is the direction of vector C then 74.7 is the angle relative to the positive X AIS if CX was negative then the vector would point to the left and we'd have to find 180° minus the 74.7 de to find the conventional angle so that's how we would solve this type of problem which involves adding two vectors and using conventional angles relative to the positive X AIS we had to really keep track of our angles and our positive and negative signs but as we'll see some Vector problems are a lot simpler so let's take a look at the next problem a person is riding a zipline between two towers that are 185 M apart when they reach the end of the zip line they are 20 M lower than when they started what is the angle of their displacement below the horizontal okay so what information are we given here a person is writing a zipline and they have some displacement between the start and end points let's draw a DOT for the starting point the two towers are 185 M apart so let's draw a horizontal line labeled 185 M It also says the end point is 20 M lower than the starting point so let's draw a vertical line labeled 20 M here's a second dot for the end point of the zipline now let's draw a displacement Vector from the start point to the end point point we could have also wed the distances like this since this question is talking about a physical scenario and it didn't mention anything about X and Y axis we're only dealing with positive horizontal and vertical distances we can draw this picture however we want we could even flip it so the displacement is going right to left as long as the end point is 185 M from the start in the horizontal Direction and 20 M down in the vertical direction if you want you could also draw the towers and the ground but we'll just use this picture so the question is asking us to find the angle of the displacement below the horizontal which means the angle between the vector and a horizontal line in the picture that's this angle for the given information it probably makes more sense to just reference the labels in the picture instead of writing out the information in words but we could write that we're trying to find Theta which is the angle so how would we find Theta well from our picture we can see the displacement and the two distances form a right triangle so we can just use the inverse tangent function to find Theta the opposite side from the angle is 20 M and the adjacent side is 185 m if we plug that in we find that theta equals 6.2 so we'd say the displacement is 6.2 de below the horizontal and that's it for problems like this where we're given some distances or lengths we just have to draw what the physical scenario might look like and use geometry to find what we're looking for if you want you could include the x and y axis and say that the X component of the displacement is 185 M and the Y component is 20 M but it's not always necessary in this case we're just dealing with positive distances and angles so using geometry is enough to answer this question now let's take a look at the next problem a plane is flying 800 kmph South with no wind over a short period of time a strong wind develops with a speed of 120 kmph in the west Direction causing the plane to drift off course assuming the pilot doesn't make any changes what is the new speed of the plane over ground and what is the angle between the plane's original Direction and its New Direction due to the wind all right there's a lot of information there let's read it again and pick out what's important the plane is flying 800 km hour south with no wind so we have a plane with a velocity vector and the plane has that velocity without any wind over a short period of time a strong wind develops with a speed of 120 kmph in the west Direction so now we also have some wind which has its own velocity Vector of 120 km perh West this causes the plane to drift off course so the direction of its motion changes the pilot doesn't make any changes which means the plane doesn't speed up or slow down and it doesn't turn to compensate for the wind or anything like that the problem is asking us to find the new speed of the plane over ground and the angle between the plane's original Direction and its New Direction so if we're looking for the new speed of the plane and its new angle that means we're looking for the magnitude and angle of the plane's new velocity Vector let's draw some things to help us solve this first we're dealing with compass directions so let's draw those for reference the plane starts with a velocity Vector of 800 kmph South so let's draw that and label it as VI for the initial velocity of the plane then later on we have some wind with a velocity of 120 kmph West the plane is still flying 800 kmph South through the air but the air happens to be moving West at 120 km/ hour so relative to the ground the plane has a new velocity Vector which is its original velocity Vector plus the velocity Vector of the wind let's label this new velocity as VF for final velocity this is like a boat pointing straight across a river but there's current pushing the boat sideways so the boat moves at an angle don't worry if this scenario isn't intuitive it should make more sense after practicing these types of problems and we'll explain this stuff more in the lesson on relative motion but for this problem we want to find the magnitude of VF and its angle relative to VI the original velocity Vector for the given information we can write down that for the plane VI is 800 km/ hour south and for the wind its velocity is 120 km/h West but like last time it might just be easier to label the pictures and we're trying to find VF the magnitude of the new velocity vector and Theta the angle between VF and VI so we're adding two vectors but since they're perpendicular to each other they already form the components of the resultant velocity Vector so all we need to do is find the magnitude and angle of the new vector and we can use these equations first let's find the magnitude if we rearrange the Pythagorean theorem equation the magnitude of V f is the hypotenuse C and A and B are the legs of the triangle which would be 800 km hour and 120 km/ hour when we calculate that we get 809 km/ hour for the magnitude of VF now let's find this angle in this case the opposite side from the angle is 120 km/ hour and the adjacent side is is 800 km/ hour when we plug those in we get 8.5 de for Theta so the new speed of the plane over ground is 809 km/ hour and the angle between the plane's original Direction and its New Direction is 8.5 de for this problem we used compass directions and we didn't use the X and Y axis we added one-dimensional vectors and found a reference angle but if the problem was asking us to add two dimensional vectors like in the first problem then we would have to keep track of positive and negative components and in that case it might make sense to use the X and Y directions to keep track of things all right let's move on to the last problem a hockey team is practicing on a rink with an imaginary X and Y coordinates system in units of meters a puck is sitting at coordinates 64 when player a passes it across the ice to player B the puck moves at a constant velocity of 16 m/s assuming no friction at an angle of 72° relative to the x-axis what are the coordinates of player B if they receive the puck in 2 seconds so what's happening in this problem we're working with coordinates a velocity vector and there is time involved we have a hockey puck that starts at some initial position then it moves with a constant velocity Vector for a period of time and ends up at a final position let's read through this again draw a picture and write down the important information a hockey team is practicing on a rink with an imaginary x and y coordinate system in in units of meters so the problem is talking about a real world scenario but we're going to describe things as if they're happening on an x and y coordinate system let's draw the X and Y axis to start the picture next a puck is sitting at coordinates 64 let's draw a DOT to represent the puck and label its coordinates since the coordinates are in meters we can write down that the initial X position is 6 M and the initial y position is 4 M player a passes it across the ice to player B the puck moves at a constant velocity of 16 m/s at an angle of 72° relative to the xais so let's draw the velocity Vector of the puck and write down the magnitude V is 16 m/s and the angle Theta is 72° the problem is asking us to find the coordinates of player B who receives the puck in 2 seconds so the problem is basically saying if the puck has this initial position then moves at this velocity what's the position of the puck after 2 seconds so let's draw a second dot for the final position of the puck where player B is and write down that we're trying to find and its coordinates X final and Y final since we're working with vectors we need the usual equations to find the components or the magnitude and angle but since this problem involves velocity time and initial and final positions let's write down these two equations for displacement and velocity they both use x but we know the equations are the same for the y direction we know the initial positions the velocity and the time so we can find the displacements and the final positions so before we dive in let's talk through how we could solve this one way is to use the velocity to find the displacement Vector of the puck during those two seconds and then find the X and Y displacements which we can add to the Puck's initial positions to find the final positions another way is to find the X and Y velocity components and use those to find the X and Y displacements and then add those to the initial positions let's try both ways to see what they look like first let's find the displacement vector and then find the components of the displacement the puck is moving 16 m/ second in a straight line at 72° so after 2 seconds the puck will move some distance in that same direction so the displacement Vector will have the same angle as the velocity Vector but what's the magnitude of the displacement since we're only looking at the motion in a straight line we can treat it as a linear motion problem velocity equals the displacement divided the period of time instead of Delta X let's use D for the magnitude of the displacement in the direction it's moving so so V equal D / time we can rearrange that to isolate D on the left and then plug in 16 m/s for V and 2 seconds for delta T that gives us 32 M for the displacement that's pretty simple math the puck moves 16 m/s for 2 seconds so it travels 32 m in that direction now that we have have D we can find the X and Y components how far the puck moved in the X and Y directions using the trig functions Delta x = d * the cosine of theta and Delta y = d * the S of theta we plug in 32 M for D and 72° for Theta that gives us 9.9 M for Delta X and 30.4 M for Delta y now we can find the final positions of the puck Delta X and Delta y equal the final positions minus the initial positions we can rearrange those to isolate the final positions which means we're just adding the initial position plus the displacement we plug in 6 M for X initial and 9.9 M for Del Delta X to get 15.9 M for xfinal then we can plug in 4 M for y initial and 30.4 M for Delta y to get 34.4 M for y final putting those together as coordinates the final position of the puck would be 15.9 comma 34.4 now let's look at the other way to solve this problem first we'll find the X and Y velocity components then we'll use those to find Delta X and Delta y so the X component VX would be V * the cosine of theta and VY is V * the S of theta we plug in 16 m/s for V and 72° for Theta and we get 4.94 m/s for V VX and 15.22 m/s for VY so as the puck moves from the initial point to the final point it moves to the right at 4.94 m/s and moves up at 15.22 m/s now we can use those to find Delta X and Delta y we could use these two equations and solve this in two steps but if we take the equation for velocity and replace Delta x with X final minus X initial and then we rearrange things we get this equation and we can calculate the final positions in one step we plug in 6 M for X initial 4.94 m/s for VX and 2 seconds for T that gives us 15.9 M for xfinal for the Y position we plug in 4 M for y initial 15.22 m/s for VY and 2 seconds for T that gives us 34.4 M for y final that's the same answer we got before we just did the steps in a different order so that's an example of a problem that combines coordinates vectors and some kinematic equations all right all right that's it for this video thanks for watching and I'll see you in the next one