What is the Maximum Reaction Time to Avoid Hitting a Deer?

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SUMMARY

The maximum reaction time for a motorist traveling at 18.0 m/s to avoid hitting a deer 38.0 m ahead, with a maximum negative acceleration of -4.50 m/s², is 0.11 seconds. If the motorist's actual reaction time is 0.300 seconds, the vehicle will be traveling at 3.69 m/s upon impact. The calculations involve using the equations of motion: Vf = Vi + at and d = Vit + 0.5at². The analysis highlights the importance of accurately assessing reaction times and braking distances in collision scenarios.

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  • Study the kinematic equations in detail, particularly Vf = Vi + at and d = Vit + 0.5at²
  • Learn about the implications of reaction time on vehicle stopping distances
  • Research the effects of different deceleration rates on collision outcomes
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UrbanXrisis
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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?
 
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It seems alright--- I am very very tired-- just getting back from soccer but looks good to me
 
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
 
thank you very much
 
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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