Indefinite Integration problem

QuarkCharmer
Messages
1,049
Reaction score
3

Homework Statement


A particle is moving with the given data. Find the position of the particle.
a(t)=t^{2}-4t+6,
s(0)=0,
s(1)=20

Homework Equations



The Attempt at a Solution


a(t)=t^{2}-4t+6, s(0)=0, s(1)=20
v(t)=\int t^{2}-4t+6 dt
v(t)=\frac{t^{3}}{3}-2t^{2}+6t+C_{1}
Then I suppose I take the antiderivative again to get to s (distance), I can't solve for the constant yet.

s(t)=\int \frac{t^{3}}{3}-2t^{2}+6t+C_{1} dt
s(t)=\frac{t^{4}}{12}-\frac{2t^{3}}{3}+3t^{2}+C_{1}t+C_{2}

So now I have two constants in there? Would I just use the parameters s(0)=0 and s(1)=20 to try to solve this like a system of 2 equations with two unknowns? I don't really know how to proceed. The other problems like this at least gave me a value for the first derivative of the original function so I could sort of work backwards.
 
Last edited:
Physics news on Phys.org
QuarkCharmer said:
So now I have two constants in there? Would I just use the parameters s(0)=0 and s(1)=20 to try to solve this like a system of 2 equations with two unknowns?

This is exactly what you do. Be more confident! :smile:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top