Question 1084556: A car slows to a stop with an average acceleration of -3.o m/s2. How long will it take to slow the car from 15m/s to a position of rest? Answer by natolino_2017(77) (Show Source):
You can put this solution on YOUR website! V(t) = V(0)+a*t
In this case V(0)= 15m/s and a = -3 m/s^2.
So V(t) = 15 -3t.
V(t) = 0 when 15-3t=0
t=5 seconds.
@natolino_
*********************Bonus**************************************
if you want the distance travelled, you use the next formula.
(Vfinal)^2 = (Vinitial)^2 + 2*a*d, when d is the distance travelled.
Vfinal = 0; Vinitial= 15 m/s ; a = -3 m/s^2.
So d =(15)^2/(2*3) = 37.5 meters.
the distance is travelled is 5 seconds if the car slow down ( if the car doesn't slow down the distance travelled in the same time would be 15*5 = 75 meters)
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