 Hello and welcome to the session. In this session we will discuss a question which says that a cool ship is travelling due north at 20 knots that is at 20 nautical miles per hour. Then part A is if the wind is blowing due east at 5 knots, what is the resultive velocity and direction of the ship and B part is if the wind velocity doubles, what is the resultive path and velocity of the ship. Now let us start with the solution of the given question. Now in the question it is given that a cool ship is travelling due north at 20 knots and in A part it is given that the wind is blowing due east at 5 knots. So we first draw the diagram representing the situation so that we can make a clear idea about the direction. Now we have these directions that is north, east, south and west. Using these directions we will draw the graph. Now we take coordinate plane and positive y axis is in north direction, positive x axis is in east direction. So we draw the initial path of the ship due north. Now the ship is travelling due north at 20 knots, a vector representing the path lies on the positive y axis 20 units long and it is given that the wind is blowing due east at 5 knots. So a vector representing the wind will be parallel to the positive x axis and it will be 5 units long. Now using the triangle rule we will find the resulted path which can be represented by a vector from the initial point of the vector representing the ship to the terminal point of the vector representing the wind. So this vector represents the resultant. Now by joining these we can see we are getting the right angle triangle at these points be o, a, b. So o, a, b is the right angle triangle. Now here you can see magnitude of vector o, a is equal to 20. Now let this magnitude be a. So a is equal to 20. Similarly magnitude of vector a, b is equal to 5. Let us denote it by b. So b is equal to 5 and let the magnitude of the resultant vector that is the vector o, b is equal to c. Now by Pythagorean theorem we have c square is equal to a square plus b square. So this applies c square is equal to 20 square plus 5 square which applies c square is equal to 400 plus 25. This means c square is equal to 425 and this applies c is equal to square root of 425 which implies c is approximately equal to 20.6 thus the resultant speed of the ship is approximately equal to 20.6 knots. Now we have to find direction of the ship and for this we will use tangent ratio. Now here this is the angle theta and this is the right angle. Now here side a, b of this triangle is perpendicular and side o, a is the base. So tan theta is equal to perpendicular upon base. Now magnitude of vector a, b is 5 and magnitude of vector o, a is 20. So tan theta is equal to 5 upon 20. This implies theta is equal to tan inverse of 5 upon 20. Now using calculator this implies theta is approximately equal to 14.0. So the resultant direction of the ship is about 14 degrees east of due north. Therefore the resultant vector is 20.6 knots at 14 degrees east of due north. Now in the second part if velocity of wind doubles then we have to find the resultant path and velocity of the ship. Now here we will use scalar multiplication to find the magnitude of the vector for wind velocity. Now it is given that velocity of wind doubles. So here scalar k is equal to 2. Now previous wind velocity was given to us as 5 knots. So magnitude of this velocity is equal to 5. Now the new wind velocity will be 10 times magnitude of vector v that is 2 into 5 which is equal to 10. So new wind velocity is equal to 10 knots. Now again we represent this situation in a diagram. Now the ship is moving with velocity 20 knots and it is already represented by this vector which is 20 units long. And now the new wind velocity that is the wind is going due east at 10 halves and it is represented by this vector which is 10 units long. Now again we will use the triangle rule. So here the resultant path can be represented by a vector from the initial point of the vector representing the ship to the terminal point of the vector representing the wind. So this is the resultant. Now here again we have got a right angle triangle. Now the magnitude of one side say A is equal to 20, magnitude of the other side say B is equal to 10 and that magnitude of this resultant vector BC. Now again we will use the Pythagorean theorem and we have C square is equal to A square plus B square which implies C square is equal to 20 square plus 10 square which implies C square is equal to 400 plus 100 and this gives C square is equal to 500 which implies C is equal to square root of 500 and we have C is approximately equal to 22.4. Now let this angle be theta and this is the right angle. Now we have to find the direction of ship. So we will use the tangent ratio. Now 10 theta is equal to perpendicular upon these that is 10 upon 20 which implies 10 theta is equal to 1 upon 2 and this gives theta is equal to 10 inverse of 1 upon 2 and by using calculator theta is approximately equal to 26.6 degrees. Now here we have obtained C is approximately equal to 22.4 which means the resultant velocity of the ship is about 22.4 knots and theta is equal to 26.6 degrees which means the resultant direction of the ship is about 26.6 degrees east of due north. It means if the wind velocity doubles the ship cruises along the path approximately 26.6 degrees east of due north at about 22.4 knots. So this is the solution of the given question and that's all for this session. Hope you all have enjoyed the session.