# Relation between the time, velocity and acceleration

This is question from I.E. Iredov's General Physics:

$1.22$ : The velocity of a particle moving in the positive direction of the $x-axis$ varies as $v = α \sqrt x$, where $α$ is a positive constant. Assuming that at the moment $t = 0$ the particle was located at the point $x = 0$.

(a) Find the time dependence of the velocity and the acceleration of the particle.

My effort:

$x=0 , t=0$

If I have $\dfrac{dx}{dt}=\alpha \sqrt{x}$,

$\int_0^x\alpha \sqrt{x}$ gives the displacement. How would I possibly bring in the time $t$ in here.

• Hi Inceptio. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. – Qmechanic May 8 '13 at 9:22
• @Qmechanic: Sorry. I wasn't aware of this. This is not the same as Math.SE. – Inceptio May 8 '13 at 9:23

Well integration is the basic method. But some observation can help too:

$$v=\alpha\sqrt x$$ $$\text{Squaring both sides:}$$ $$v^2=\alpha^2 x$$

We know $v^2\propto x$ gives constant acceleration.

$\text{Remember} :v^2=u^2+2as.$

So, comparing it with this equation we get $$v^2=0+2\frac{\alpha^2} 2 x$$

So, acceleration =$\alpha^2/2$ and velocity =$at=\alpha^2 t/2$

Differentiate $\dfrac{dx}{dt}= \alpha \sqrt{x}$ with respect to time again to get:

$$\dfrac{d^2x}{dt^2} = \dfrac{\alpha}{2 \sqrt{x}} \dfrac{dx}{dt} = \dfrac{\alpha}{2 \sqrt{x}} \alpha \sqrt{x} = \dfrac{\alpha ^2}{2}$$

Then:

$$\dfrac{dx}{dt} = \dfrac{\alpha ^2 t}{2}$$

and:

$$x = \dfrac{(\alpha t)^2}{4}$$

for the given initial conditions.