Re: was Kinematics of Throwing Balls; really braking distances

[Larry Woolf <larry.woolf@GAT.COM>]

A solution:

reaction distance  = vt

stopping distance = v^2/(2a)

So v^2/(2a) + vt = d

a=5.18m/s^2

t=1.6 s

d=48.78m

Use the quadratic formula to solve

v^2/(2a) + vt - d = 0

for v.

Larry Woolf;General Atomics;6995 Flanders Dr.;MS 78-107;San Diego CA
92121-2975; Ph:858-526-8575;FAX:858-526-8568; www.ga.com; www.sci-ed-ga.org

-----Original Message-----
From:  Julie Hilsenteger
Sent: Thursday, September 26, 2002 2:40 PM
To: PHYS-L@lists.nau.edu

OK, I cannot seem to get it.  I know that for the reaction time part the car
is going constant velocity and covers a distant dependent on the velocity.
Then the car brakes and ends with a final velocity of zero.  But all I know
is the total distance the car is allowed to go, the acceleration, the final
velocity.  The initial velocity for the braking part is the velocity for the
reaction time part and the distance traveled braking is the 48.78m -
distance
traveled during the reaction time part.  I seem to have too many unknowns.
So, how do you solve it?