Deceleration (Driving a Car-Help)

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The discussion revolves around calculating the stopping distance of a car when encountering a deer on the road. The initial calculations using the equation d = v0*t + 1/2*a*t^2 were incorrect due to misunderstanding the reaction time and stopping process. The correct approach involves recognizing that the car travels for 0.5 seconds before braking begins, covering an additional distance. Participants suggest using the equation v_f^2 - v_i^2 = 2*a*d to determine the stopping distance accurately. The conversation emphasizes the importance of correctly applying physics equations to solve the problem effectively.
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Deceleration (Driving a Car--Help)

You're driving down the highway late one night at 20m/s when a deer steps onto the road 35m in front of you. Your reaction time before stepping on the brakes is .5s, and the maximum deceleration of your car is 10m/s.

A. How much distance is between you and the deer when you come to a stop?
B. What is the maximum speed you could have and still not hit the deer?

So I don't think I started the problem correctly...
I used the equation d=v0*t+1/2a*t^2
so then d=20*.5-1/2*10*.5^2
d=8.75

so then I subtracted 8.75m from 35m and got 26.25m

It was wrong, then I thought 8.75 was it but that was wrong so I don't know what I did wrong. Could someone tell what I did wrong. I think it might of been the equation so I look for a problem like that in my book but couldn't find anything. Please someone help. Just the equation don't ruin the fun :redface:
 
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It does not take .5 seconds for the car to stop. You drive for another .5 seconds (or 10m) until you begin to stop. Since you know the speed and deceleration of the car, now use another equation: v_f^2 - v_i^2 =2 a d.
 
d=distance right? but then
vf-vi=2ad
0-20=2*-10*d
-20/-20=1=d
so d doesn't have any sign so it doesn't make since...
 
sorry being an idiot thank you it work just fine
 
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