Chapter 5: Exercises with Solutions

Section 5.5MotionFor the distance traveled, x 0 = 0 ft since the car hasn’t traveled any distance at timet = 0 s.x = x 0 + v 0 t + 1 2 at2= 0 ft + (88 ft s × 10 s) + (1 2 × −3ft s 2 × (10 s)2 )= 0 ft + (88 ft s × 10 s) + (1 2 × −3ft s 2 × 100 s2 )= 0 ft + (88 ft s × 10 s) + (1 2 × −3ft s 2 × 100 s2 )= 0 ft + 880 ft − 150 ft= 730 ft33. We are given the initial velocity v 0 = 180 ft/s and the initial position–the objectis shot from the surface of earth, so its initial distance is x 0 = 0 ft. It is well knownthat the acceleration due to gravity is a = −32 ft/s 2 when we measure in feet (and it isnegative because it is pulling down on the object). The maximum height of the objectoccurs when the velocity is v = 0 ft/s. So our task is to find x. First, find t by using,v = v 0 + atv − v 0 = atand plugging in the known values:at = v − v 0t = v − v 0at = v − v 0a0 ft/s − 180 ft/s=−32(ft/s)/s= −180ft/s ×= 5.625 sAnd now use the distance formula to find x.x = x 0 + v 0 t + 1 2 at21s−32ft/s= 0 ft + (180 ft s × 5.625 s) + (1 2 × −32ft s 2 × (5.625 s)2 )= 0 ft + (180 ft s × 5.625 s) + (1 2 × −32ft s 2 × 31.640625 s2 )= 0 ft + (180 ft s × 5.625 s) + (1 2 × −32ft s 2 × 31.640625 s2 )Version: Fall 2007