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virtus
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
Solve the following: \[\ln(t^2 + 3) = 5\] \[t^2 + 3 - e^5 = 0 \Longrightarrow t = \pm \sqrt{e^5 - 3} \]
If 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: \[P-a = \ln (t^2 + 3)\]
callum 29 how to i go from where you left me at?
what is the answer?
I have to go now, later guys :)
I have to go now, later guys :)
Well \[e^5\] 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.
I 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 re-wind if you like. It tells you it takes 12.18 seconds to get there and 12.18 seconds to come back...