|
Title: Deceleration Post by: hboardz133 on Apr 2, 2012 Car A is traveling at twice the speed of car B. They both hit the brakes at the same time and undergo identical decelerations. How does the distance required for car A to stop compare with that for car B?
Title: Re: Deceleration Post by: knight3434 on Apr 2, 2012 The stopping distance of car A would be four time of Car B.
Ex. Use Vf^2=vi^2 +2ad lets say the deceleration is -2m/s car A has speed of 6 0=(6)^2+2(-2)(d) d=9 Car B has speed of 3 0=(3)^2+2(-2)(d) d=2.25 9/2.25 =4 Title: Re: Deceleration Post by: subro on Apr 2, 2012 If Car A is driving twice as fast, meaning it's velocity is 2x(Car B), then I would conclude that it would take twice as long to decelerate to a stop (velocity)
so I would say 1/2 times Car A's stopping distance. |