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

Question

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 · Answer

Solve the following:
$$\ln(t^2 + 3) = 5$$
$$t^2 + 3 - e^5 = 0 \Longrightarrow t = \pm \sqrt{e^5 - 3} $$

anonymous · Answer

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:
$$P-a = \ln (t^2 + 3)$$

anonymous · Answer

didn't*

anonymous · Answer

callum 29 how to i go from where you left me at?

anonymous · Answer

*do i go

anonymous · Answer

what is the answer?

anonymous · Answer

12.0587 seconds?

anonymous · Answer

I have to go now, later guys :)

anonymous · Answer

I have to go now, later guys :)

anonymous · Answer

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.

anonymous · Answer

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...