Question

The displacement of a particle is zero at t=0 seconds and it is x at t seconds. it stars in a positive x direction with a velocity that varies v=k√x where k is a constant . Show that velocity varies with time.

Answer 1

The displacement of a particle is zero at $t=0$ seconds and it is $x$ at $t$ seconds. It stars in a positive $x$ direction with a velocity that varies $v=k\sqrt{x}$ where $k$ is a constant. Show that velocity varies with time.

By definition:

$$v = \frac{dx}{dt}$$

x – displacement

t – time

And:

$$v = k\sqrt{x} \quad \Rightarrow \quad x = \left(\frac{v}{k}\right)^2$$

or

$$dv = \frac{k}{2\sqrt{x}} dx$$

Therefore:

$$dx = \frac{2\sqrt{x}}{k} dv = \left| x = \left(\frac{v}{k}\right)^2 \right| = \frac{2v}{k^2} dv$$

Substitute to $v = \frac{dx}{dt}$:

$$v = \frac{\frac{2v}{k^2} dv}{dt}$$

or:

$$\frac{\frac{2v}{k^2} dv}{v} = dt$$ $$\frac{2}{k^2} dv = dt$$

Integrate:

$$\frac{2v}{k^2} + C = t$$

C – some constant

displacement of a particle is zero at $t=0$ seconds and $v = k\sqrt{x}$ therefore $v=0$ then $t=0$:

$$C = 0$$ $$\frac{2v}{k^2} = t$$ $$v = \frac{k^2}{2} t$$

Answer: $v = \frac{k^2}{2} t$