Physics Problem

[Silmarian]

What acceleration does a rocket need to reach a speed of 210 m/s at a height of 1.2 km?

How would I go about solving this?  

[VorpalNeko]

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Say you have some acceleration a(t). The area under the graph represents the change in velocity. If the acceleration is constant (horizontal graph), you just get a rectangle of area at. Hence, from rest, that's v = at.

Doing this again for the velocity graph gives you change in position. For the line v = at, you get a triangle of area (1/2)(at)(t). So, from zero initial position, you travel x = (1/2)at².

You're almost guaranteed to have seen both of these relationships before. Here's where we get slightly creative. If we have the time it takes to get to that point, we'll have out answer. So all you have to do is notice that
x = (1/2)at² = (1/2)(at)(t) = (1/2)vt
lets you solve for t, since you are given x and v. Then you can just take a = v/t = v²/(2x).

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The net work done on the payload (mass m), is given by max, where ma is the net force on it and x is the distance, Hence max = (1/2)mv², or a = v²/(2x). Note that the gravitational force is included in the net force.

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We can also abandon the physically unreasonable assumption of a constant acceleration, but then the problem is not sufficiently well-defined to give a single numerical answer.  

[bigmanlittlepackage]

Displacement (S)= 1200 metres, Initial velocity(U)=0m/s, end velocity=210m/s(V), acceleration=what we want to find out(A), Time= non applicable(T).

The equation we can use is V^2=U^2+2*A*S. U is zero so this cancels down to V^2=2*A*S. This makes 210^2=2(A*1200). This can rearrange to make 22050=A*1200, then 22050 divided by 1200 is 18.375. So the answer should be 18.375ms^(-2)  

[Mecill]

VorpalNeko, I thought the area under the graph was velocity, rather than change in velocity?  

[VorpalNeko]

No, it's change in velocity. It can't be velocity because the initial velocity is not determined by acceleration. You're probably confusing it with "rate of change in velocity", which would indeed make that statement wrong, as that's just acceleration.  

[Mecill]

Oh, ok. Thanks! So in general am I correct to say if you were explaining this with calculus, the derivative is the rate of change of a variable (a being the derivative of v) and the definite integral is the change in a variable from the initial condititon (change in v being the definite integral of a)? I forgot the FOTC. sweatdrop Thanks again.