Deceleration and reaction time physics problem

[xgoddess210]

Homework Statement

You're driving down the highway late one night at 16 m/s when a deer steps onto the road 39 m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 12 m/s^2 . How much distance is between you and the deer when you come to a stop? What is the maximum speed you could have and still not hit the deer?

Homework Equations

reaction time is 1/2 of 16m so it is 8m
x=vot*.5at

The Attempt at a Solution

t=1.33 s

I know there are 20 meters between me and the deer because I was able to figure out the first part of the question with help from my brother. I can't remember the calculations because I left my physics book in my room and I'm in the library currently.

I know we tried two different ways to solve the problem and got wrong answers. Honestly, I don't even know how to start the second half of the problem. Please give me an outline on what to do, at least, so I can figure this problem out. Thank you!

 

[Hootenanny]

Homework Statement

You're driving down the highway late one night at 16 m/s when a deer steps onto the road 39 m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 12 m/s^2 . How much distance is between you and the deer when you come to a stop? What is the maximum speed you could have and still not hit the deer?

Homework Equations

reaction time is 1/2 of 16m so it is 8m

This is good. So when you hit the brakes, what is the distance between you and the deer?

x=vot*.5at

This is perhaps, not the best equation to use. Write down all the variables you know and the one that you want to know. Which kinematic equations contains *only* the variables that you have listed?

 

[xgoddess210]

I've already solved the first part of the problem. There are 20 meters between me and the deer. I'm just not sure about the other part. What is the maximum speed you could have and still not hit the deer?

 

[Hootenanny]

I've already solved the first part of the problem. There are 20 meters between me and the deer. I'm just not sure about the other part. What is the maximum speed you could have and still not hit the deer?

Sorry, I thought you were still have problems with the first part.

For the second part, how far will you stop from the deer if you *just* avoid hitting it?

 

[xgoddess210]

I understand what the problem is saying, I just don't know how to find the answer. I don't know what equation to use or how to even go about solving it.

 

[Hootenanny]

I understand what the problem is saying, I just don't know how to find the answer. I don't know what equation to use or how to even go about solving it.

Perhaps if you answered my question and used my hints you might be able to give it a go:

For the second part, how far will you stop from the deer if you *just* avoid hitting it?

This is perhaps, not the best equation to use. Write down all the variables you know and the one that you want to know. Which kinematic equations contains *only* the variables that you have listed?

 

[Duncan1382]

Homework Statement

You're driving down the highway late one night at 16 m/s when a deer steps onto the road 39 m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 12 m/s^2 . How much distance is between you and the deer when you come to a stop? What is the maximum speed you could have and still not hit the deer?

Okay, I'll walk you through it in words and see if you can solve it with numbers...

1) Subtract the distance that you "react" from the total distance between you and the deer to get to the point when you start to "decelerate" (tell your teacher that's a bad word -- use "negative acceleration"). Since you have constant velocity, use the equation x = vt.

2) Next, determine the distance it takes to stop. You do NOT have time, so you cannot use the normal delta x equation. Use the equation WITHOUT time. Solve for x. Use the maximum acceleration rate. You've got to make acceleration negative!

v squared = v initial squared plus (2 times acceleration times delta x)
v2 = vo2 + 2ax

3) Subtract that distance from the distance when you started accelerating and you've got your answer.Sum Up: 39m - (distance to react) - (distance to stop) = answer---
EDIT: Sorry, didn't' even see part two...

1) Again, still use the equation without time, but this time we are solving for v initial.

Just plug in the following values and solve:
vo = ?
vf = 0 m/s
delta x = distance from when you started reacting (found in first problem)
a = -12 m/s^2

Don't forget that positives and negatives will mess you up if you don't use them correctly!

 

[xgoddess210]

Ok. Maybe I'm doing calculations wrong, but I got 27 and MasteringPhysics says that is wrong. Here is what I did:

v2=vo2+2ax

0=vo2+(2*-12)(31) 31 is 39-8, and .5*16=8

0=vo2+(-744)

squared root of 744=Vo

Vo=27

Where did I go wrong?

 

[xgoddess210]

help?