+249 When a car's brakes are applied, it travels 5 feet less in each second than the previous second until it comes to a complete stop. A car goes 45 feet in the first second after the brakes are applied. How many feet does the car travel from the time the brakes are applied to the time the car stops?
+37146 Given a = -5 ft/sec^2
and xf = 45 and t = 1 find vo
xf = xo + vot + 1/2 a t^2 xo = 0
45 = vo (1) + 1/2 (-5) 1^2 yields vo = 47.5 ft/s
Find TIME to stop given vo = 47.5 ft / s a = -5 ft/s^2 and vf=0 (the car has stopped)
vf = vo + a t
0 = 47.5 - 5 t yields t = 9.5 s
Find DISTANCE to stop given vo=47.5 ft/s a = -5 ft/s^2 and t= 9.5 s
xf=xo + vo t + 1/2 a t^2 xo=initial position being measured from = 0
xf = 47.5 (9.5) + 1/2 (-5)(9.5^2) Yields xf = 225.625 ft to stop
