![[Two-Dimensional Motion]](/23718/media/2dmotion.JPG)
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Review the basics of vectors. In life, situations aren't always just dropping rocks from the roof of a building or pushing a block across a table at constant speed (that would be boring). Life takes on a more 3-dimensional aspect (although there are actually 4 dimensions), but let's take it at a comfortable pace and consider motion in 2 dimensions. When it is said that an object is moving in a "plane" it usually means that it's moving on a flat (2-D) coordinate axis like the x and y axes (i.e. as if it were moving on a smooth flat surface) So...in what kind of situation would an object move in 2-D motion, for example? Well, in 1-D motion, the object moved in exclusively one direction, either a horizontal "x" or vertical "y" direction. So in 2-D, it's not either or; the object moves in both the x and y directions. Recall some equations for velocity and acceleration:
In considering these types of physics problems, you'll treat the x and y motions as two different cases (i.e. considering only one direction at a time). To illustrate, glance over this sample problem: A certain airplane, which is diving at an angle of 30º below the horizontal, is flying at a speed of 205 miles per hour. A bomb is 102.5released at point A. The pilot hits the target (civilian) at point B which is 3.0 mi away. (a) How long was the bomb in the air? (b) At what altitude was the bomb released? <figure> Solution: Ok so the plane is going at 205 miles per hour when the bomb is released. Therefore, this is the speed at which the bomb is moving when it is released at point A. Find the x and y components of the velocity: <pic of 30 degrees right triangle> vy = 205(sin30º) mi/h = 102.5 mi/h vx = 205(cos30º) mi/h = 177.5 mi/h x = vit + at2; you're given that x = 3.0 mi, and you know that vx = 177.5 mi/hr and a = 0 (horizontally), so solve for t. The equation reduces to x = vxt t = x/vx = (3.0 mi)/(177.5 mi/h) = 0.0169 h = 1.7 x 10-2 h Now we want to find the height of point A. If it took t hrs for the bomb to hit the ground, we can use the equation that involves height and time (in terms of the y-direction, otherwise known as "freefall") which is nearly the same equation as in the earlier part of the question: y = vyt + at2; a = acceleration due to gravity and solve for y. y = (102.5 mi/h)(0.0169 h) + (9.8 m/s2)(0.0169)2 = 1.735 mi = 1.7 mi
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