Question

The speed of a particle on the x axis, $\mathit{x}\mathbf{\ge }\mathbf{0}$, is always numerically equal to the square root of its displacement x. If $\mathit{x}\mathbf{=}\mathbf{0}$when $\mathit{t}\mathbf{=}\mathbf{0}$, find x as a function of t. Show that the given conditions are satisfied if the particle remains at the origin for any arbitrary length of time ${\mathit{t}}_{\mathbf{0}}$and then moves away; find x for $\mathit{t}\mathbf{>}{\mathit{t}}_{\mathbf{0}}$for this case.

Short answer

Answer

At $x=t=0$, the displacement $x\left(t\right)=\frac{t^2}{4}$.

At $t>t_0$, the displacement $\left(t\right)=\frac{{\left(t-t_0\right)}^2}{4}$.

Step-by-step solution

Given information

It is given that the speed of a particle on the x axis, $x\ge 0$, is always numerically equal to the square root of its displacement x

Definition of differential equation

A differential equation is an equation that contains at least onederivative of an unknown function, either an ordinary derivative or a partial derivative.

Solve differential equation

Speed is defined as the first time derivative of the displacement. So the differential equation is as follows:

$$\begin{gathered} \frac{dx}{dt}=\sqrt{x} \\ \int \frac{dx}{\sqrt{x}}=\int dt \\ 2\sqrt{x}=1t+t_0 \end{gathered}$$

Here, $t_0$is the integration constant.

Therefore, the general solution is

$x\left(t\right)={\left(\frac{t+t_0}{2}\right)}^2$.

For the initial condition, when $x=0,t_0=0$

Therefore, $x\left(t\right)=\frac{t^2}{4}$.

For the case $t>t_0$, replace t with $t-t_0$in the above equation to get

$x\left(t\right)=\frac{{\left(t-t_0\right)}^2}{4}$.

Therefore,

at $x=t=0$, the displacement $x\left(t\right)=\frac{t^2}{4}$.

at $t>t_0$, the displacement $x\left(t\right)=\frac{{\left(t-t_0\right)}^2}{4}$.