Sunday, October 28, 2007

Re: [Math4u] mathematicaly challenged

I agree with the equation of motion.

Part of what is asked is

"find the quadratic function based on the stopping distance for 55 mph
and 65 mph and the fact that at 0 mph the stopping distance is 0 feet."

I think we can give an equation for this with [D65] and [D55] in the equation without knowing the values for the stopping distances [D65] and [D55].

Part of what is asked is

"use that equation to predict the stopping distances for 55 mph and 65
mph."

You are correct; we cannot predict actual stopping distances without values for [D55] and [D55]. What is silly is that the stopping distances for 55 and 65 mph will be the same numbers that we started with, a good check of the formula.

Regards,

Brian


 
----- Original Message -----
Sent: Sunday, October 28, 2007 7:34 AM
Subject: Re: [Math4u] mathematicaly challenged

Newtons equation of motion for stopping distance from speed
s = vt + (0.5)at^2
a is equal to the rate of change of velocity, or acceleration (in this case it will be negative acceleration), t is the time to stop, v the velocity.
As you can very well see
The data that they gave you is insufficient


 
On 10/28/07, Brian E. Jensen <brianejensen@prodigy.net > wrote:

Stopping distance is proportional to the square of the speed. It must be because stooping time is proportional to speed. Average speed = (s+0)/2=s/2.

Stopping distance = stopping time * average velocity (s/a)*(s/2)=s^2/(2a)

d=constant*s^2

There is always a reaction time plus an amount of time to move the foot from the gas to the brake.

d=constant*s^2 + s * reaction time

d=s^2/(2a)+reaction time * s

Looks like a quadratic equation to me.

Since the problem doesn't mention reaction time, it is assumed to be zero.

If we know the stopping distance for either 55 or 65 mph, then we can calculate a. Otherwise we can give the formula with a variable for the stopping distance at 65 mph as the answer.

Regards,

Brian

----- Original Message -----
Sent: Saturday, October 27, 2007 12:37 PM
Subject: Re: [Math4u] mathematicaly challenged

 

Forgot the graph, sorry

On 10/27/07, Douglas Anderson <djandersonza@gmail.com > wrote:
See the attachment. This is a graphical representation of what you are after.
Plotting stopping distance vs speed is a simple linear equation, but you do need to know what your rate of deceleration is. As it stands it can be anything you like.
For arguments I have assumed that the breaking force will be the same for both cars, hence the faster car will travel a greater distance.
Hence
if
s = speed (mph)
d = breaking distance (ft)
d = a s (where a would be the deceleration constant)
 
I do not think they have given you enough information

 
On 10/27/07, Brian E. Jensen <brianejensen@prodigy.net > wrote:

Theoretically, the deceleration is constant. We have no information to the contrary,

D = distance from stopping point

A = deceleration, constant

D=A*T^2/2

V=A*T

Honestly, I know this is solvable but don't know where this is going. I just keep fearlessly writing down what I know and we'll discover together where this goes.

The question doesn't say what it is looking for, but I am forced to guess that they want the equation of stopping distance verses velocity since these are the only two things mentioned in the question.

Stopping distance = feet

Velocity = feet / second = mph*5280/3600 = mph*528/360 = mph*132/90 = mph*66/45= mph*22/15

A=V/T=2*D/T^2 from the above 2 equations

V=2*D/T

T=2*D/V

This looks confusing because I still don't know exactly how we are gong to get the answer. But we should get the same answer by using either the stopping distance for 65mph or the stopping distance for 55mph.

Let [D65] = the stopping distance for 65 mph.

Let [T65] = the stopping time for 65 mph

Now I am getting stuck and confused, so what I decide is to derive the relationship between stopping distance and velocity.

Let's use intuition:

Stopping time is proportional to velocity

T=[C1]*V

Stopping distance = Stopping time * Average velocity

Stopping distance = Stopping time * velocity/2

Stopping distance = [C1]*V * velocity / 2

Stopping distance = [C1]*[mph]*(22/15) * [mph]*(22/15)/2

Stopping distance = [C1]*(22/15)^2*(1/2)*[mph]^2

Stopping distance = [C2]*[mph]^2

[C2]=[D65]/65^2

Our formula will be:

D=[C2]*V^2

Or

D={[D65]/65^2}*V^2 answer

Check

[D55] should equal {[D65]/65^2}*55^2

If not, check to see if [D65]/[D55]=(65/55)^2

If not, then the problem may be flawed.

Regards,

Brian

----- Original Message -----
Sent: Friday, October 26, 2007 6:53 PM
Subject: Re: [Math4u] mathematicaly challenged

 

You would need a little more information than what they have given, but just look up Newtons equations of motion for some inspiration.

On 10/26/07, caglepieniazekgladys < caglepieniazekgladys@yahoo.com> wrote:

I am trying to complete the following homework assisnment.

find the quadratic function based on the stopping distance for 55 mph
and 65 mph and the fact that at 0 mph the stopping distance is 0 feet.

use that equation to predict the stopping distances for 55 mph and 65
mph.

I have tried to follow the examples in the book, but nothing seems to
work.

can anyone help.

Thanks

Elasha





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