Kinematics - One Dimension Problem

[antimatter1422]

The problem is stated as follows:

A typical automobile has a maximum deceleration of about 7m/s^2; the typical reaction time to engage the brakes is 0.50 s. A school board sets the speed limit in a school zone to meet the condition that all cars should be able to stop in a distance of 4 m.
a) what maximum speed should be allowed for a typical automobile? (Answer = 10.6 m/hr)
b) what fraction of the 4 m is due to the reaction time (Answer = 2.38 m)

* Now i know that part b will be a piece of cake if i get part a. But whenever i do part a (its been several tries...) i get around 13.6 miles per hour, not 10.6.

* Also, it really irks me that i can't figure out these kinematics problems without any help. The book they come from is " Paul A. Tipler physics for scientists and engineers." This book shows the answers, but it doesn't show how they got the answers, so i can't learn how to do them.

* In conclusion, if anyone could point me in the right direction, i'd appreciate it. Also, does anyone know how to learn this stuff so i can do it on my own. I can draw out the situation, so i think i understand it conceptually. Is algebra skills holding me back (like setting the problem up, maybe that's where I am messing up). so should i just try more problems, may b easier so i can work my way up and do problems like these? or go back to the basic concepts?
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What I've got so far is as follows:

The first .5 s of the 4 m there is no acceleration (reaction time). So the distance of the first part of the 4 m traveled (lets say "x") is Vo/2. (From X = Xo + Vavg*t). The second part is where the acceleration is, and you are going to end up at rest. So the distance of the 2nd part of the 4m (let's say "4-x") is (Vo/2)*(t-.5s). If i re arrange this equation i get x = 4 - ((2Vo*t - Vo)/4). So i can set this x equal to Vo/2. I can simplify to get Vo = 16/(2t+3) , or t = (16 - 3*Vo)/(2Vo). If i go back to where t = .5 s and plug that in, i get Vo = 4 m/s. If i change 4m/s to mi/h, i get 8.948 mi/hr. NOT the right answer. (and I am not sure what i did different, but last time i got 13.67 mi/hr)

- What am i doing wrong?

 

[antimatter1422]

CORRECTION :
Answer to part a) is not 10.6 m/hr. It is 10.6mi /hr

 

[moose]

Why does it go from using meters, to using miles?

 

[antimatter1422]

i guess to make the problem more realistic., because they are talking about the school zone speed limit (in the usa). but u should be able to just convert ur m/s in the end to mi/hr.

 

[antimatter1422]

moose, if you worked it out, what did you get for you final answer to part a?

 

[Fermat]

You error was in treating t as a variable rather than as a constant.
t is the time of the total journey and is fixed.
The eqn "Vo = 16/(2t+3)" is true only for a fixed value of t, not a variable one.

I found it easier to use the suvat eqn,

v² = u² - 2as

and treating it as two separate movements - one at constant speed (for 0.5 s) and the other at constant deceleration.

I got the right answer :)

 

[H_man]

So the distance of the 2nd part of the 4m (let's say "4-x") is (Vo/2)*(t-.5s).

Okay if x is the dist. traveled before braking then$$x = \frac {1}{2} V_o$$

So dist. left to brake => 4 - x = 4 - $$\frac {1}{2} V_o$$

If you then substitute this distance into your kinematic equations you should have some joy.

BTW, it helps when reading what people have posted if they don't squeeze all their calculations into one big paragraph.

Hope this helps

 

[antimatter1422]

thanks

thanks everyone for your help. i got it.
ill remember the paragraph bit for next time :tongue: