 Hello and welcome to the session. In this session we will solve problems involving velocity and other quantities that can be represented by vectors. Let us start with magnitude and direction of vectors. Now suppose we have position vector v in the coordinate plane with initial point at origin, having components given by our vector x, y. So, x is horizontal component and y is vertical component like theta with a positive angle between the x axis and the vector v measured in anti-clockwise direction. Now this angle theta will give us direction of vector v. Now see, this is the right angle triangle with high potential use, magnitude of vector v like this triangle will be triangle o, b, c. So, o c is equal to magnitude of vector v. Thus, direction of the vector is given by tan theta is equal to perpendicular upon base. Now perpendicular is y and base is x. So, tan theta is equal to y upon x where x is not equal to 0. And applying path of motion in this triangle we have magnitude of vector v is equal to square root of x square plus y square. Now if we do the angle theta then we can find the horizontal and the vertical components of the vector using trigonometric ratios. Now we know that sin theta is equal to perpendicular upon high potential use. So, in this triangle we have sin theta is equal to y upon magnitude of v and from this we obtain the vertical distance y is equal to magnitude of v into sin theta. And cos theta is equal to x upon magnitude of v from this we obtain the horizontal distance x that is equal to magnitude of v into cos theta. So, coordinates of this terminal point c will be magnitude of v cos theta that is magnitude of v into cos theta and magnitude of v into sin theta. Thus the horizontal component x is equal to magnitude of v into cos theta and vertical component y is equal to magnitude of v into sin theta. Now before starting velocity problems of vector let us understand meaning of true theory. Now in displacement problems direction of a vector needs to be found. Now one way to measure the direction of travel is to use true theory. Now true theory is the measurement of a clockwise angle from the true north direction. Now let us understand velocity vector and its components. Now a vector that represents the direction and speed of an object in motion is called velocity vector. For example an air draft is flying on a bearing of 65 degrees at 500 miles per hour find the components of the velocity of the airplane. Now first of all let us draw its diagram. Now we have four directions north, south, east and west. Now we have given that the plane is flying at a bearing of 65 degrees and we know that the true bearing is the measurement of a clockwise angle from the true north direction. So from north direction moving in clockwise direction by 65 degrees we get vector v which is the velocity vector. Now the angle it makes with horizontal axis is 90 degrees minus 65 degrees that is equal to 25 degrees. Here in direction horizontal component of velocity will be in east direction and vertical component is in north direction. Let v be the velocity of the airplane. A bearing of 65 degrees is equivalent to a direction angle of 25 degrees. Now according to western formula is magnitude of v so magnitude of v is equal to 500. Now we know that horizontal component x is equal to magnitude of v into cos theta and vertical component y is equal to magnitude of v into sin theta. Now here we have theta is equal to 25 degrees and magnitude of v is equal to 500. So horizontal component of vector v is equal to 500 into cos of 25 degrees and vertical component of vector v is equal to 500 into sin of 25 degrees. So the velocity components are given by the ordered pair 500 into cos 25 degrees 500 into sin 25 degrees which is approximately equal to the ordered pair 153.15, 211.31. Now the components of the velocity give the eastward and northward speeds that is airplane travels about 453.15 miles per hour eastward and about 211.31 miles per hour northward as it travels at 500 miles per hour on the bearing of 65 degrees. Now let us discuss force components. Now we can use vectors in finding the components of a force exerted on a body. Now let us discuss an example. A volume is being pulled by a rope that makes a 25 degrees angle with the ground. The person is pulling with the force of 103 newtons along the rope, determine the horizontal and vertical components of the vector. Now here we want to find the components of 103 newtons vector so that he simply need to draw a triangle and then solve for the unknown sides the triangle will be drawn using the force vector from above along with horizontal and vertical components. Now see this diagram we have a wagon and a rope at an angle of 25 degrees with horizontal line that is ground. So this will be the horizontal component of force which is denoted by fx and this will be the vertical component of force that is denoted by fy. Now we have vector f whose magnitude is 103 newtons so that magnitude of vector f is equal to 103 and angle theta is equal to 25 degrees. So using component form of vector we have horizontal component of force that is fx is equal to magnitude of f into cos theta and vertical component fy is equal to magnitude of f into sin theta. So fx is equal to 103 into cos of 25 degrees which is equal to 93.34 and fy is equal to 103 into sin of 25 degrees which is equal to 43.5 thus horizontal component of force that is fx is approximately equal to 93 newtons and vertical component of force that is fy is approximately equal to 44 newtons. Now if a right hander triangle is not signed then we can make use of laws of sign and cosine in finding the magnitude and direction. So in this session we have discussed how to solve problems involving velocity and other quantities that can be represented by vectors and this completes our session. Hope you all have enjoyed the session.