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!