Net force as a function of time?

[Comet1592]

Homework Statement

I need to find the Net force as a function of time given x and y as functions of time

Relevant Equations

Given that:
y(x)=(2h(x/L)^3)+3h(x/L)2
x(t)=(v_xo)t-L
y(t)=h(((2(v_x0)t/L)^3)-3(v_xo)t/L)^2)+1)
m=300,000Kg
v_x0=100mi/hr
L=25miles
h=10miles

All I've done so far is think about F_net. Since F=ma, and a is a vector, I was thinking that I should find the x and y components of a and then try to calculate F_net that way, but I'm confused as to where I should use x(t) and y(t). Or instead, thinking about it as the change in momentum over time as delta(mv)/delta(t). Honestly, I have no clue how to figure this out, just some general physics knowledge from calc-based general physics 1.

 

[RPinPA]

Yes, $F_{net} = ma$

That can be interpreted as a vector equation, relating $\vec{F}_{net}$ to $\vec a$.

Now, what is the definition of $\vec a$ in terms of the position vector?

 

[Comet1592]

You mean the second integral of the position function x?

 

[Orodruin]

You mean the second integral of the position function x?

No. Acceleration is not the second integral of position.

 

[Comet1592]

Oops! Its the second derivative of the position function then?

 

[Orodruin]

Indeed. So the force is?

 

[Comet1592]

So then that would be the second derivative of y(x)? y''(x)=((12hx)/L^3)+((6h)/L^2)
Then I plug in the numbers for L and h, and get a function in terms of x. From there all I have to do is multiply by mass to get net force? But its still in terms of x and not time... Do I just substitute t in at that point? I just don't get how it's related to x(t) and y(t).

 

[Comet1592]

Oh wait... do I plug x(t) into my y(x) equation after I get the second derivative? That would give me a function in terms of t...

 

[RPinPA]

Oops! Its the second derivative of the position function then?

Right. $\vec a$ is the second derivative of $\vec x$ with respect to time.

So then that would be the second derivative of y(x)? y''(x)=((12hx)/L^3)+((6h)/L^2)

No, because (a) that's not the position function $\vec x(t)$ and (b) that is not a time derivative.

That is one of the components of position, and you took its derivative with respect to x.

The position is a vector. It has in this case two components, each of which is varying in time. $\vec x(t) = (x(t), y(t))$.

Now, what does it mean to take the derivative of that thing with respect to time? What do you get? Forget your particular problem for the moment. What does it mean in general to take the time derivative of a vector?