# Acceleration and reaction time

1. Sep 27, 2004

### UrbanXrisis

A motorist is traveling at 18.0 m/s when he sees a deer in the road 38.0 m ahead.

(a) If the maximum negative acceleartion of the vehicle is -4.50 m/s^2, what is the maximum reation time t of the motorist that will allow him to avoid hitting the deer?

thought process...
Vf=Vi+at
0m/s=18m/s+(-4.5m/s^2)t
t=4

d=Vit+.5at^2
d=18(4)+.5(-4.5)(4)^2
d=36m

That means there is 2 meters left for a crash.
d=Vit+.5at^2
2=18(t)+.5(0)t^2
t=.11s

Max reaction time: .11s

(b) If his reaction time is actually .300s, how fast will he be traveling when he hits the deer?

.3-.11=.19s
the driver goes .19 seconds over

4s-.19s=3.81s
how long it takes for him to hit to hit the deer

Vf=Vi+at
Vf=18m/s+(-4.5m/s^2)(3.18s)
Vf=3.69m/s

Is this work valid?

2. Sep 28, 2004

### Tom McCurdy

It seems alright--- I am very very tired-- just getting back from soccer but looks good to me

3. Sep 28, 2004

### Tom McCurdy

I may have time to take a better look at it in the morning I worked out the beginnign steps and got the same thing as you but my dad came down and I need to get some sleep being 1:30 I will work on it tomorrow if I get a chance... hopefully I was a little helpful

goodnight

4. Sep 28, 2004

### UrbanXrisis

thank you very much

5. Sep 28, 2004

### Leong

I don't think the work is valid. For the first case, after 0.3 s seeing the deer, the speed of the motorcycle is definitely has decreased because the brake is applied after 0.11 s. But for the second case, after 0.3 s, the speed is still 18 m/s and only at that time the brake is applied. The case is different, so you can't mix them up.

If i were you, i would find how far has the motorcycle gone after 0.3 s, and then find the distance ahead before hitting the deer and use $$v^2=u^2-2as.$$

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