To stop a car you require first a certain reaction time to begin breaking; then the car slows under a constant

Question

To stop a car you require first a certain reaction time to begin breaking; then the car slows under a constant breaking deceleration. Suppose that the total distance moved by your car during these two phases is 186 ft when its initial speed is 50mi/h, and 80 ft when the initial speed is 30 mi/h. What are a) your reaction time and b) the magnitude of the deceleration.

If you can answer this you are a genius.

Thanks in advance

Answer 1

Assuming the same deceleration magnitude, you calculate using
d(t)=v0*(t-tr)-.5*a*(t-tr)^2
and
v(t)=v0-a*(t-tr)

since you have two conditions, you can solve.

1 mph = 1.47 ft/s
at 50 mi/hr
186=50*1.47*tr+
50*1.47*(t1-tr)-
.5*a*(t1-tr)^2

simplify
186=50*1.47*t1-
.5*a*(t1-tr)^2

0=50*1.47-a*(t1-tr)

use
50*1.47=a(t1-tr)
(73.5+a*tr)/a=t1
186=73.5*tr+73.5*(t1-tr)-
.5*a*(t1-tr)^2

186=73.5*t1-.5*a*(t1-tr)^2

186=73.5*(73.5+a*tr)/a-
.5*a*((73.5+a*tr)/a-tr)^2


at 30 mph
80=30*1.47*t2-
.5*a*(t2-tr)^2

0=30*1.47-a*(t2-tr)
(44.1+a*tr)/a=t2

80=44.1*(44.1+a*tr)/a-
.5*a*((44.1+a*tr)/a-tr)^2

We now have 2 equations for tr and a

80=1944.81/a+44.1*tr-
.5*a*((44.1+a*tr)/a-tr)^2

186=5402.25/a+73.5*tr-
.5*a*((73.5+a*tr)/a-tr)^2

80*a=1944.81+44.1*a*tr-
.5*(44.1)^2

186*a=5402.25+73.5*a*tr-
.5*(73.5)^2

160*a=3889.62+
88.2*a*tr-1944.81

372*a=10804.5+
147*a*tr-5402.25



160*a=1944.81+88.2*a*tr
372*a=5402.25+147*a*tr

solve for a

800*a/3=3241.35+147*a*tr
372*a=5402.25+147*a*tr

316*a/3=2160.9
a=20.515

Now solve for tr

800*a/3=3241.35+147*a*tr
tr=0.739 s
Check
372*a=5402.25+147*a*tr

tr=0.739 s

checks

For fun I graphed v(t) and x(t) for both cases. See
http://i142.photobucket.com/albums/r88/odu83/stop1.jpg


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