Posted by **will** on Sunday, February 10, 2008 at 4:29pm.

A car is traveling 26 m/s when the driver sees a child standing in the road. He takes 0.8 s to react, then steps on the brakes and slows at 6.0 m/s2. How far does the car go before it stops?

- physics -
**Damon**, Sunday, February 10, 2008 at 4:51pm
In .8 seconds he goes 26 * .8 meters

then the problem is acceleration with initial speed Vo = 26 and acceleration = -6 m/s^2

time to deacceletate to 0:

v = Vo + a t

0 = 26 - 6 t

t = 26/6

distance = .8* 26 + 26 t - (1/2) 6 * t^2

- physics -
**drwls**, Sunday, February 10, 2008 at 4:55pm
The distance travelled during the "reaction time" t1 is

X1 = V*t1 = 26 m/s * 0.8 s. = ?

The time required to decelerate to 0 is t2 = V/a = 4.33 s

The distance travelled while decelerating is the average velocity multipled by t2.

X2 = (V/2)(V/a) = V^2/(2a) = ?

Add X1 and X2 for the answer.

- physics -
**Damon**, Sunday, February 10, 2008 at 4:58pm
Ah, good. I am off for a while :)

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