An object is moving along X axis with position as a function of time given by $x = x(t)$. Point $O$ is at $x = 0$. The object is definitely moving towards $O$ when
1. $\mathrm dx/\mathrm dt < 0$
2. $\mathrm dx/\mathrm dt > 0$
3. $\mathrm d(x)²/\mathrm dt < 0$.
4. $\mathrm d(x)²/\mathrm dt > 0.$

I can't understand why the correct answer is $\mathrm d(x)²/\mathrm dt < 0$ instead of $\mathrm dx/\mathrm dt < 0$. Is there a mathematical formula for knowing whether a particle is moving towards $O$ in this situation?


$x(t)$ simply means the value of the x co-ordinate at time $t$.

Draw a diagram of the x axis on which the object is moving. $x$ increases (becomes more +ve) to the right and decreases (becomes more -ve) to the left.

$dx/dt=v<0$ means that the x co-ordinate is decreasing as time passes. $x$ is becoming more -ve, the object is moving left. If it is to the right of O (ie $x>0$) then it will be moving towards O. However, if it is to the left of O ($x<0$) then the object will be moving away from O.

Likewise for $dx/dt=v > 0$. This means that $x$ is increasing, the object is moving right. If the object is left of O it will be moving towards O but if it is already to the right of O it is moving away from O.

$x^2$ represents the squared distance from O. It is always positive so it does not have the above complications with signs. If $d(x)^2/dt<0$ then the (squared) distance from O is always decreasing, whether the object is to the right or left of O. So it is moving towards O.

Likewise for $d(x)^2/dt>0$. The distance from O is increasing : the object is getting further from O, whichever side of O it is already.


$\frac{d}{dt}x(t)^2=2x(t)\frac{dx(t)}{dt}<0$ by the chain rule

Because the product $2x(t)\frac{dx(t)}{dt}$ is negative, $x(t)$ or $\frac{dx(t)}{dt}$ has to be negative.

Suppose $x(t)<0$ then the position of the particle is 'negative' i.e. it's to the left of the origin. Then the volocity, $\frac{dx(t)}{dt}$ is positive, so the particle moves to the right, towards the origin.

Suppose $x(t)>0$. The particle is to the right of the origin and it's speeds has to be negative, so the particle moves towards the origin as well.

  • $\begingroup$ So what is meant by x(t)? $\endgroup$ – pranjal verma Jun 13 '18 at 17:09
  • 1
    $\begingroup$ It's simply a function wich assigns to every time $t$ a certain position along the x-axis. What exactly is your confusion? $\endgroup$ – QuirkyTurtle98 Jun 16 '18 at 6:30
  • $\begingroup$ I can't understand if there is an increase in t will the value of x decrease or increase? $\endgroup$ – pranjal verma Jun 21 '18 at 3:12

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