A hot-air balloon is 160 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going 60 mi / hr (88 ft/ s). If the balloon rises vertically at a rate of 11 ft/s, what is the rate of change of the distance between the motorcycle and the balloon 5 seconds later? Let x be the horizontal distance from the balloon to the motorcycle, y be the vertical distance from the balloon to the road, and z be the distance between the motorcycle and the balloon. Write an equation relating x, y, and z. 2 2 =x2 + y2 Differentiate both sides of the equation with respect to t. (2x ) at + (2y) at = (22 ) The rate of change of the distance between the motorcycle and the balloon after 5 seconds is about V (Round to two decimal places as needed.)