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# Calculus/acceleration

Question
QUESTION: Hi and thanks in advance in anticipation of a relevant reply and answer to the following question.If an object is accelerating such that its acceleration from a fixed point is proportional to the inverse square of its distance from the the fixed point,how do I calculate  time as a function of distance or vice versa?    many thanks and kindest regards,Barry.

Since v(t) = s'(t), v(t) = ct^(c-1).
Since a(t) = v'(t), a(t) = c(c-1)t^(c-2).

Since it says acceleration a(t) is inversely proportional to distance s(c) squared,
that means t^(c-2) ~ 1/t^(2c).  Multiplying both sides by t^(2c) gives us
t^(3c-2) ~ 1.  This means that the exponent on the left is 0.
That is, 3c - 2 = 0, or 3c = 2, so c = 2/3.

This gives s(t) = kt^(2/3) for some constant k.
This implies that v(t) = kt^(-1/3) for some other constant k.
This implies that a(t) = kt^(-4/3) for still another constant k.

As can be seen, the power of t on the acceleration is -4/3.
As can also be seen, the power on the inverse square of distance is 2(-2/3) = -4/3.

Thus, we have the same powers, so we know that distance s(t) = kt^(2/3) for some constant k.

---------- FOLLOW-UP ----------

QUESTION: Hi, and thanks for your reply to my query,however not being a mathematician I find your answer difficult to understand.In leibniz notatation I am asking for a solution to the second order differential equation as follows....d^2x/dt^2=k/x^2.Perhaps if I ask the question in a different form it may help.If a small pebble is dropped from an arbitrary height in the Earth's inverse square gravitational field and neglecting any extraneous forced such as air resistance etc how long would it take to reach the Earth's surface for any given height from which it was dropped.What equation is required to give time as a function of distance?I didn't phrase the question initially as above because wished to avoid including "mass"in the solution.Many thank's again and kindest regards,Barry.

To solve d²x/dt² = k/x², multiply both sides by x² dt giving x² d²x = k d²t.
Integrating once gives x³/3 dx = kt + M dt, where M is a constant.
Integrating again gives x^4/12 = kt²/2 + Mt + N, where N is a constant as well.

The solution for an equation with respect to time could be found with the quadratic equation.
We have A = k/2, B = M, and C = N - x^4/12.

Actually, if it was dropped from rest, I believe that M = 0 and N is the starting height H0.
This gives x^4/12 = kt²/2 + H0.  In this case, t = sqrt[2(x^4/12 - H0)/k].

The true equation for dropping a pebble from rest at height H0 is given in metric by
d = -4.9t² + H0 since gravity is -9.8m/s².  This would mean the time would be
t = sqrt((H0 - d)/4.9).  The total time would be when d is 0, so t = sqrt(H0/4.9).

Is this an answer to the question?

Calculus

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