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
