Finding the Time to Reach the Halfway Point with Proportional Acceleration

[einstein314]

Homework Statement

A particle moving from a point A to a point B, 1 meter away, travels in a straight line in such a way so that its acceleration is proportional to the distance left to point B. If the particle arrives at point B in 1 second, how long did it take for the particle to reach the point halfway to point B?

Homework Equations

I suppose we need that the acceleration is the second-derivative of position.

The Attempt at a Solution

So we know that a(t) = \frac{d^2p}{dt^2} = k(1 - p(t)) (and a(1) = 0 and p(0) = 0 and p(1) = 1), but I don't know how to solve this differential equation. Once p(t) is found, t can be found by equating p(t) = \frac{1}{2}.

 

[TSny]

You can try writing the differential equation in terms of a new dependent variable y that is defined in terms of p. Can you see how to define y(t) in terms of p(t) so that you get a simpler differential equation?

 

[ehild]

Homework Statement

A particle moving from a point A to a point B, 1 meter away, travels in a straight line in such a way so that its acceleration is proportional to the distance left to point B. If the particle arrives at point B in 1 second, how long did it take for the particle to reach the point halfway to point B?

Homework Equations

I suppose we need that the acceleration is the second-derivative of position.

The Attempt at a Solution

So we know that a(t) = \frac{d^2p}{dt^2} = k(1 - p(t)) (and a(1) = 0 and p(0) = 0 and p(1) = 1), but I don't know how to solve this differential equation. Once p(t) is found, t can be found by equating p(t) = \frac{1}{2}.

substitute 1-p(t) by u. What equation do you get for u? Are you familiar with it?

 

[einstein314]

Define y(t) = p(t) - 1. Then \frac{d^2y}{dt^2} = \frac{d^2p}{dt^2} and the differential equation becomes:
\frac{d^2y}{dt^2} = -ky
Auxiliary equation is r^2 + k = 0 so r = \pm i \sqrt{k}. Then y = e^a(c_1 cos(bt) + c_2 sin(bt)) = e^{(0)}(c_1 cos(t\sqrt{k}) + c_2 sin(t\sqrt{k})) = c_1 cos(t\sqrt{k}) + c_2 sin(t\sqrt{k}). Then p(t) = y(t) + 1 = c_1 cos(t\sqrt{k}) + c_2 sin(t\sqrt{k}) + 1. I think I can figure the rest out myself. Thanks all!