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Velocity and Distance in 2 Dimensions

	Given: v=1.5m/s $\theta=23^\circ$ t=7.0s
	Find: distance(d) using x and y components
	$v_x=v\cdot \cos(\theta) = (1.50 m/s) \cos(23^\circ)=1.38m/s$ $v_y=v \cdot \sin(\theta) = (1.50 m/s)\cdot \sin(23^ \circ)=0.59m/s$
d= $r \cdot t$ $d_x =v_x \cdot t = (1.38m/s)(7.0s)=9.67m$ $d_y=v_y \cdot t = (0.59 m/s)(7.0 s) = 4.10m$
	$d^2 = (9.67m)^2+(4.10m)^2$ $d=10.5m$