Anti-derivative motion problem

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Homework Help Overview

The problem involves determining the constant deceleration required for a motorcycle to brake from 30 mph (44 ft/sec) to a complete stop over a distance of 45 feet, without using physics formulas.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formulation of the velocity function and question the correctness of the initial attempt. There is exploration of the relationship between acceleration and the position function, with some participants suggesting different forms of the velocity equation.

Discussion Status

The discussion is ongoing, with participants providing insights into the relationships between variables and questioning assumptions about the equations used. Some guidance has been offered regarding the definitions of variables and the implications of the equations presented.

Contextual Notes

Participants are restricted from using physics formulas, which influences their approach to the problem. There is also a focus on ensuring that the final position is accurately defined within the context of the problem.

htoor9
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Homework Statement



The State cycle program requires motorcycle riders to be able to brake from 30 mph(44ft/sec) to 0 in 45 feet. What constant deceleration does it take to do that? No physics formulas allowed.


Homework Equations



Integral rules and whatnot

The Attempt at a Solution



I think v(t) = 44t + V0 but I don't know. I'm just really stumped
 
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htoor9 said:

Homework Statement



The State cycle program requires motorcycle riders to be able to brake from 30 mph(44ft/sec) to 0 in 45 feet. What constant deceleration does it take to do that? No physics formulas allowed.


Homework Equations



Integral rules and whatnot

The Attempt at a Solution



I think v(t) = 44t + V0 but I don't know. I'm just really stumped
If the equation is v(t) = 44t + v_0, the motorcyclist will be speeding up, not slowing down.

What you have is
[tex]\frac{d^2 s}{dt^2} = - a[/tex]
where a is a positive constant, and v(0) = 44 (ft/sec). You can define s(0) however you want, just so that s(final) is 45 feet away.
 
Mark44 said:
If the equation is v(t) = 44t + v_0, the motorcyclist will be speeding up, not slowing down.

What you have is
[tex]\frac{d^2 s}{dt^2} = - a[/tex]
where a is a positive constant, and v(0) = 44 (ft/sec). You can define s(0) however you want, just so that s(final) is 45 feet away.

so if the derivative is going to be -a then v(t) = -44t + v0?
 
Does that give you 44 ft/sec if you plug in t = 0?
What is v0?
 
CompuChip said:
Does that give you 44 ft/sec if you plug in t = 0?
What is v0?

Ok so it would be v(t) = -at + 44...anyways I ahead and tried to solve the problem. Could anyone check my solution? He said to leave it in fraction form and I got a = 1936/90 or 968/45.
 
anyone? does 21.5 deceleration make sense as an answer?
 

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