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swin2013
Group Title
A particle starts at x(0) = 2. If its velocity is given by v(t) = ln(1+t), find its position at t = 5
 one year ago
 one year ago
swin2013 Group Title
A particle starts at x(0) = 2. If its velocity is given by v(t) = ln(1+t), find its position at t = 5
 one year ago
 one year ago

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campbell_st Group TitleBest ResponseYou've already chosen the best response.1
the primitive of v(t) will give the displacement equation x(t) so find the indefinite integral of \[\int\limits \ln(1 + t) dt\] you also know that when t = 0, x(t) = 2 so the particle starts 2 units to the right of the origin. you will need to use these to evaluate the constant in the indefinite integral. Lastly, when you have your equation, substitute t = 5 and evaluate to find the postiion from the origin
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
by u substitution correct? i think i am doing that process wrong
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
well its a standard integral \[\int\limits \ln(ax + b) dx = \frac{(ax + b)}{a} \ln(ax + b)  x + c\]
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
oh that is where i get confused. lol. because i did: y = lnu u = 1+t dy/du = ulnu du/dx = 1 dy/dx = dy/du * du/dx and then i add that to the constant 2.
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix can you help me? loll maybe i'm thinking waaayy too much into this
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
you need to use integration by parts so f = ln(u) dg = du df = 1/u du g = u so then you have \[uln(u)  \int\limits 1 du\] which gives the integral of uln(u)  u and when u = t + 1 you get  t + (t + 1)(ln (t + 1)  1) or (t + 1)log(t +1)  1
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
AHHH ok thanks :)
 one year ago
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