Here is a quick example of how trigonometric relationships can allow us to find the x and y components of any vector, as long as the angle of the vector, relative to the axis system, is known. For that we use simple right-triangle trigonometry, and, if necessary, non-right angle trig. Calculate the actual speed of the boat (in knots) and its deviation (in degrees) from its intended courseįinally, and more often, we need to resolve a vector of known magnitude that forms an angle with the x- or y-axes that is not a 30˚, 45˚ or 60˚ angle. It travels across a lake on a day with a 5-knot cross wind (perpendicular to the intended direction of travel) from the right. On a windless day, a boat can travel at a maximum forward speed of 10 knots. Calculate the magnitude of the component velocity vectors along the ground and straight up (x and y, respectively). After how many miles will the slower car be 1 hour behind the faster?Ī projectile is fired upward at an angle of 45˚ with an initial velocity of 30 m/s.
Two cars begin a trip together, one traveling at 65 mi./h, and the other at 75 mi./h on the same freeway. How fast can an airplane, with a maximum speed of 160 Km/h in still air, travel 250 miles if it has a steady headwind of 15 Km/h ? These vector velocity problems can be solved using basic vector knowledge and the special triangles (no trigonometry required).
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