Re: [Math4u] mathematicaly challenged

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 -----

From: Douglas Anderson

To: Math4u@yahoogroups.com

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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