Graphical Relationship Between ∆x and v

[tenbee]

Homework Statement

You're pushing a car up an inclined plane (a hill). Which graph matches the relationship between ∆x and v?

Homework Equations

W = K.E. + P.E.
thus: F∆xcos$\theta$ = 1/2mv² + mgh

The Attempt at a Solution

The answer is C. I thought the answer was D - I don't understand why the relationship is a negative. Please share an explanation with me.

 

[darkxponent]

yes answer is c

Ax= Acos(theeta)

so (delta)x = 1/2 * Ax*t^2

and v= A * t

two equations give

delta(x) = 1/2 * cos(theta) * v^2 * 1/A

gives a prabola on x-axis

 

[tenbee]

yes answer is c

Ax= Acos(theeta)

so (delta)x = 1/2 * Ax*t^2

and v= A * t

two equations give

delta(x) = 1/2 * cos(theta) * v^2 * 1/A

gives a prabola on x-axis

Thanks darkxponent! Could you tell me which component in the equation F∆x = 1/2mv² I could specifically look at and determine the graph is a negative exponent function? Is it because there's a "1/2"? If that is true... will any coefficient < 1 result in a negative exponential function?

 

[cupid.callin]

Thanks darkxponent! Could you tell me which component in the equation F∆x = 1/2mv² I could specifically look at and determine the graph is a negative exponent function? Is it because there's a "1/2"? If that is true... will any coefficient < 1 result in a negative exponential function?

its not exponential but a parabolic function

y=Ax² is an eqn of parabola with y as its axis.
more on parabolas: http://en.wikipedia.org/wiki/Parabola

 

[darkxponent]

Could you tell me which component in the equation F∆x = 1/2mv² I could specifically look at and determine the graph is a negative exponent function? Is it because there's a "1/2"? If that is true... will any coefficient < 1 result in a negative exponential function?

by using diffrential calculus u can find the shape of the graph

if diffrentiation>0 implies graph increeasing
if diff<0 inmplies graph decreasing

and if double diff>0 :graph up shaped { y= x^2}
if double diff<0 :graph down shaped {y = - x^4}


magnitude of coefficient has nothing to do with the shape of graph
but + or - sing changes the shape of the graph only to create mirror image of the same

 

[tenbee]

by using diffrential calculus u can find the shape of the graph

if diffrentiation>0 implies graph increeasing
if diff<0 inmplies graph decreasing

and if double diff>0 :graph up shaped { y= x^2}
if double diff<0 :graph down shaped {y = - x^4}magnitude of coefficient has nothing to do with the shape of graph
but + or - sing changes the shape of the graph only to create mirror image of the same

Okay, ignoring all else where is the negative component in this equation F∆xcos$\theta$ = 1/2mv^2??

Here's the practice problem verbatim: A student applies force to a stalled car over a distance ∆x to increase its kinetic energy. Which graph BEST represents the relationship between the car's speed and the pushing distance?

Please don't use calculus in the explanation - I'll be clueless.

 

[darkxponent]

Please don't use calculus in the explanation - I'll be clueless.

u can't even think of physics without calculus!

 

[tenbee]

u can't even think of physics without calculus!

Yes you can - just memorize the equations : ) Besides I wasn't an engineering or physics major in college and took physics without calculus as electives.

 

[tenbee]

Update:

1) The correct graph is a square root function, so it looks like the inverse of an exponential function. ...parabola... hyperbole... sigmoidal... etc... none-of that mattered because I just wanted to relate the mathematical function of the graph to the equation I gave, and not so much the graph's shape and how it moves n such.

2) This is a square root function because when you solve for v in the equation F∆xcosθ = 1/2mv² + mgh ... we get $\sqrt{}$[2(F∆xcosθ - mgh)/m] = v

So basically I stopped one step ahead of finishing the problem... the function they were asking for (per the graphs) was v(∆x) not v²(∆x)! Otherwise I would have gotten the question correct

Thanks for your help!