# Chapter 4: Kinematics in 2D  Motion in a plane, vertical or horizontal  But, the motion in the x- and y-directions are independent, except that they.

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Chapter 4: Kinematics in 2D  Motion in a plane, vertical or horizontal  But, the motion in the x- and y-directions are independent, except that they are coupled by the time  Therefore, we can break the problem into x and y ``parts’’  We must use vectors: displacement r = x + y velocity v = v x + v y acceleration a = a x + a y  Usually, y x r x y

Two Sets of Kinematic Equations We can solve problems using the same methods as for 1D, but now we need to consider both x and y components simultaneously

Example: Motorcycle Jump Consider a motorcycle jumping between two buildings separated by a distance  x. The difference in heights of the buildings is  y. What initial velocity must the motorcycle have to just make it to the other building? What is the time to cross to the other building? What is the final velocity on impact? xx yy x y v0v0 v 0x = ?, v 0y = 0 a x = 0, a y = -g

 The x-component v x of the velocity remains constant throughout the flight time, v x =v 0x, - we neglect air resistance - a x = 0 - therefore, nothing to affect x-motion  Because of gravity, once the motorcycle is in the air, its speed in the y-direction v y increases from zero, points down, and therefore the height decreases  The magnitude of the resultant velocity also increases and the angle of the resultant velocity vector (with respect to the x-axis) changes v 0x v 1x = v 0x V 2x= v 0x v1v1 v 1y v 2y v2v2 22 11

 X-direction motion is the same as if motion occurred on a flat surface  Y-direction motion is equivalent to dropping the motorcycle What about 2D motion in the horizontal plane?  No acceleration due to gravity  Can rotate coordinate system to reduce problem to 1D - if the motion is in a straight line  If motion has a curvature - must be treated as 2D (topic of Chapter 6) x z x’ z’ v

Return to Motorcycle Problem x-direction 1.  x = (v 0x +v x )/2  t = v 0x  t   t =  x/v 0x 2.Same information as first equation 3. v x =v 0x since a x =0 4. Same information as third equation y-direction 1.  y = (v y  t)/2  v y = 2  y/  t 2.  y = -g  t 2 /2   t =  (-2  y/g) 3. v y = -g  t 4. v 2 y = -2g  y *three different ways to get v y

 What is the final velocity when motorcycle lands on the other roof? We know from the 3 rd x-direction equation that: v x = v 0x Therefore we need only v y. From 4th y-direction equation:

If  y  0 or g  0 or v 0x , then v  v 0x If  y  or g  or v 0x  0, then v  v y Let’s add some numbers:  x = 50.0 ft,  y = -20.0 ft, g=32.2 ft/s 2 v 0x =  x  (-g/(2  y))=(50.0 ft)  ((-32.2 ft/s 2 )/(2(-20.0 ft))) = 44.9 ft/s  t =  x/v 0x = (50.0 ft)/(44.9 ft/s) = 1.11 s Or  t =  (-2  y/g) =  (-2(-20.0 ft))/(32.2 ft/s 2 )) = 1.11 s v y = -g  t = -(32.2 ft/s 2 )(1.11s) = -35.9 ft/s Or v y = -  (-2g  y) = -  (-2(32.2 ft/s 2 )(-20.0 ft)) = -35.9 ft/s

=  [(44.9 ft/s) 2 +(35.9 ft/s) 2 ] = 57.5 ft/s  = sin -1 (v y /v) = sin -1 (-35.9/57.5) = -38.6  v 0x vyvy v  y x v = 57.5 ft/s @ - 38.6 

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