Chapter 5: Exercises with Solutions

Section 5.5Motion= 0 ft + 542.08 ft − 270.948 ft≈ 271.1 ftSo, it takes the car approximately 6.72 s to come to a stop, and it travels about 271 ftduring this time period.37. The object is released from rest, so v 0 = 0 m/s. t = 5 s and we are asked tofind how far the object will fall and what its velocity v will be after those 5 s. Thewell-known value for the acceleration due to gravity is a = −9.8 m/s 2 when using themetric system (and it is negative because it pulls the object downward).First, let’s find v.v = v 0 + at= 0 m (s + −9.8 m/s )× 5 ss= 0 m s − 49m s= −49 m/sNext we find the distance x that the object falls. We assume the object is high enoughabove the earth’s surface that it will not hit the ground within 5 s and cut the fallshort. And we know that x 0 = 0 m, as the object hasn’t fallen any distance when t = 0s.x = x 0 + v 0 t + 1 2 at2= 0 m + (0 m s × 5 s) + (1 2 × −9.8m s 2 × (5 s)2 )= 0 m + (0 m s × 5 s) + (1 2 × −9.8m s 2 × 25 s2 )= 0 m + (0 m s × 5 s) + (1 2 × −9.8m s 2 × 25 s2 )= 0 m + 0 m − 122.5 m= −122.5 mThe negative value indicates that the height of the object decreased (it fell).So, the object drops with a velocity of −49 m/s, falling 122.5 m.39. The object is released from rest, so v 0 = 0 ft/s; its initial height is x 0 = 352 ft.Its only acceleration is due to gravity, which is well known to be a = −32 ft/s 2 whenusing ft (negative because it is pulling down, decreasing the height of the object). Tofind how long it will take to hit the ground, we must find the time t when x = 0 (theheight is zero, or the object is on the ground).We want to find t, so solve the formula for t.Version: Fall 2007