v = Vi + a t
v = 30 - 3.5 t
when does it stop ?
t = 30/3.5 = 8.57 s
average speed during stop = 15 m/s
so
15 *8.57 = 129 meters, heavens to murgatroid, ran over the buoy
so how fast was he going when he hit the buoy>
d = 100 = 30 t - (1/2) 3.50 t^2
100 = 30 t - 1.75 t^2
t = 4.5 or 12.6 seconds
use 4.5 seconds, 12.6 is after you run over it and come back and hit it again in reverse
then v = 30 - 3.5*4.5 = 30 - 15.75= 14.25 m/s as he runs over the buoy
A speedboat moving at 30 m/s approaches a no-wake buoy marker 100 m ahead. The pilot
slows the boat with a constant acceleration of 3.50 m/s2
by reducing the throttle.
1 answer
1 answer