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Two particles start moving along the same straight line starting at the same moment from the same point in the same direction. The first moves with constant velocity $u$ and the second with constant acceleration $a$. During the time that elapses before second catches the first, the greatest distance between the particles is ?


I was able to get to the answer using a simple mathematical approach and considering relative motion:

Let particle with constant velocity be 1 and particle with constant acceleration be 2.

$u_{(1,2)}$(Initial velocity of 1 with respect to 2) $=u$

$a_{(1,2)}=-a$

let $x$ be the distance between the particles at any time $t$,

$x=ut-\frac{1}{2}at^2$

Using maxima-minima on this, (Taking derivative and equating to 0) , we get $t=\frac{u}{a}$

Putting this in the equation, we get $x=\frac{u^2}{2a}$


But when I looked more on these values, I found that $t=\frac{u}{a}$ happens to be the time when the velocity of the 2nd particle becomes $u$ that is, equal to the constant velocity of the first.

I don't think this is just a mere mathematical coincidence and this must mean something but I'm not able to understand this.
Does the greatest distance between two particles like these always occur when the velocity of the 2nd becomes equal to the 1st? If yes, then why? Can someone please help me understand this

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    $\begingroup$ Hi Prajwal. It's a great question and one you can probably answer for yourself with a hint. Suppose we view the motion in the rest frame of the non-accelerating particle. In this frame the accelerating particle is initially moving away at velocity -v. Consider what the motion looks like in this frame and you'll have your answer :-) $\endgroup$ Sep 24 at 4:22
  • $\begingroup$ Also note that if you put $u=at$ into the equation for $x$ you get $$x=at^2-\frac12 at^2$$ $\endgroup$
    – PM 2Ring
    Sep 24 at 4:23
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    $\begingroup$ @JohnRennie Thanks for that hint! I believe looking from the frame of the non accelerating particle, we would first see some separation happening till the point $t=\frac{u}{a}$ then we would see approach, hence maximum distance occurs exactly at that point $\endgroup$ Sep 24 at 5:10
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    $\begingroup$ Yes :-) It's like throwing a stone upwards. The maximum separation between you and the stone is at its apex when it is stationary relative to you i.e. travelling at the same velocity as you. $\endgroup$ Sep 24 at 5:27
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    $\begingroup$ To put it more directly: when the velocities are equal, the accelerating particle stops dropping behind the constant v particle and starts catching up. $\endgroup$ Sep 24 at 18:34
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Using a $v-t$ graph is always helpful.

vtgraph

Here black line is the velocity of the particle with uniform velocity(1) and the red line is the velocity of the particle moving with uniform acceleration(2).

As you know, the area of $v-t$ graph gives the displacement. If both particles starts at the same point, after time $t_1$, when velocities of both particles become equal, first particle is ahead of the second particle by the distance represented by blue area. Aftermath, the second particle starts the chasing. And at time $t_2$, it is able to cover the initial gap, so they meet again.

Therefore until $t_1$ they increase the gap and then until $t_2$ they decrease the gap. Thus at the time when the two particles have the same velocity, there is the maximum gap.

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The particles have positions given by $$x_1=ut \\ x_2=\frac{1}{2}at^2$$ the first for constant velocity and the other constant acceleration. To find the elapsed time when they are both at the same location, you can set $x_1=x_2$ which will give a time $$t=\frac{2u}{a}$$ where both objects coincide.

But (as has been pointed out in the comments by John Rennie and yourself) if you were in the frame of the first particle, looking at the second (accelerating) particle, it will move away initially, then after some time $\frac{t}{2}$ it stops momentarily (at which point both particles do have the same velocity), and then it begins to move toward to the first. That is, half way at $$\frac{1}{2}\frac{2u}{a}=\frac{u}{a}$$ is the time where the distance between the two is the greatest.

This also corresponds to $$x=\left(\frac{u^2}{a}-\frac{u^2}{2a}\right)=\frac{u^2}{2a}$$ from the equation $$x=ut-\frac{1}{2}at^2$$ Setting the derivative of this equation equal to zero gives the same value for $t(=\frac{u}{a})$. Note that $$v=\frac{dx}{dt}=u-at$$ and so setting that equal to zero corresponds to the same time. So it is not at all a coincidence.

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No it is not a coincidence at all as we know that for $\vert x_2-x_1 \vert$ to be as maxima $\frac{d}{dt}\vert x_2-x_1 \vert=0$ using the extremum theorem which is basically saying $\vert v_2-v_1 \vert=0$ you will use this property various times for problem solving that maximum seperation occurs when the relative velocity is $0$.

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