How Fast Can You Drive Without Hitting the Deer?

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To determine the maximum speed without hitting the deer, the equations of motion are applied considering the distance of 35 meters, a reaction time of 0.5 seconds, and a maximum deceleration of 10 m/s². The initial speed (Vi) must be calculated by setting the stopping distance equal to 35 meters, factoring in the time to react. The relationship between initial speed and time is established as Vi = 10T, allowing for substitution into the distance equation. Solving these equations reveals the maximum speed that can be maintained without collision. The calculations indicate that the maximum speed should be adjusted based on the time taken to stop after the reaction delay.
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Homework Statement


you're driving down the highway late one night at 20ms when a deep steps onto the road 35m infront of you. your reaction time before steeping on the brakes is 0.5s and the max deceleration of ur car is 10m/s^2

what is the max speed you could have and still not hit the deer?

deltaX=35m V=?
a=-10m/s^2 reaction time = 0.5s


Homework Equations



Xf= Xi+ViT+1/2aT^2
Vf-Vi=aT
Vf^2=Vi^2+1/2a(deltaX)

The Attempt at a Solution


so this is what i tried

i pluged the numbers into the first equation to get

35=0+ViT+1/2(-10)T^2

and i have 2 missing variables so i tried to get it from the other equation

Vf-Vi=aT
0-Vi=(-10)T then that didnt work so i tried the other equation

Vf^2=Vi^2+1/2a(deltaX)
0=Vi^2+1/2(-10)35
Vi= 13.23m/s

and i know that Vi can't be right because i did the question before it and it had Vi of 20m/s and it was 5m from hitting the deer...i don't know what i did wrong can someone help please?
 
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Jennifer001 said:

Homework Statement


you're driving down the highway late one night at 20ms when a deep steps onto the road 35m infront of you. your reaction time before steeping on the brakes is 0.5s and the max deceleration of ur car is 10m/s^2

what is the max speed you could have and still not hit the deer?

deltaX=35m V=?
a=-10m/s^2 reaction time = 0.5s


Homework Equations



Xf= Xi+ViT+1/2aT^2
Vf-Vi=aT
Vf^2=Vi^2+1/2a(deltaX)

The Attempt at a Solution


so this is what i tried

i pluged the numbers into the first equation to get

35=0+ViT+1/2(-10)T^2

and i have 2 missing variables so i tried to get it from the other equation

Vf-Vi=aT
0-Vi=(-10)T then that didnt work so i tried the other equation

Vf^2=Vi^2+1/2a(deltaX)
0=Vi^2+1/2(-10)35
Vi= 13.23m/s

and i know that Vi can't be right because i did the question before it and it had Vi of 20m/s and it was 5m from hitting the deer...i don't know what i did wrong can someone help please?
In order to JUST miss hitting the deer, it must take all 35 meters to slow to 0 m/s. So you have TWO of the equations you just gave: 35=0+ViT+1/2(-10)T^2 for the 35 m and 0-Vi=(-10)T for the time to slow to 0. Now you have two equations in the two "unknown" numbers Vi and T. 0-Vi= -10T tells you that Vi= 10T. Replace Vi by that in the first equation to get a simple equation for T alone. Once you have found T, put that into Vi= 10T to find Vi.

 
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