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anonymous
 4 years ago
if displacement is given by x=ln(t^2 +3) where x is in cm and t is in seconds, find the exact time when the particle is 5cm to the right of the origin
anonymous
 4 years ago
if displacement is given by x=ln(t^2 +3) where x is in cm and t is in seconds, find the exact time when the particle is 5cm to the right of the origin

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Solve the following: \[\ln(t^2 + 3) = 5\] \[t^2 + 3  e^5 = 0 \Longrightarrow t = \pm \sqrt{e^5  3} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If it did't start at the origin, say it started at the point a where\[P>a>0\] and where \[P\] is a distance of interest, then the exact time it would take for the particle to reach P having started at a is calculate by solving: \[Pa = \ln (t^2 + 3)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0callum 29 how to i go from where you left me at?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have to go now, later guys :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have to go now, later guys :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well \[e^5\] is a constant, approximately equal to 148.413 so we have two solutions: \[t = \pm \sqrt{148.413  3} = \pm \sqrt{145.413} = \pm 12.18249 = \pm 12.18\] to 2 d.p. Obviously, there are two solutions, however only one makes sense. You cannot have a negative value for time  otherwise you are going back in time! So you jut take the positive one.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think the negative time corresponds to the time taken by the particle to move from the point 5cm away from the origin back to the origin, in rewind if you like. It tells you it takes 12.18 seconds to get there and 12.18 seconds to come back...
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