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$x_1 (t) = 2t +1$ and $y_1 (t) = 4t^2$

$x_2(t) = 3t$ and $y_2 (t) = 3t$

How to calculate whether $x$ and $y$ is collide?

Or, in which way I can calculate this? (I do not need the actual answer as I can do calculation by myself.)

Thank you

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2 Answers

up vote 3 down vote accepted

You want to find whether there exists value of $t$ for which, simultaneously, $x_1(t)=x_2(t)$ and $y_1(t)=y_2(t)$.

So you want to figure out if the equations $$\begin{align*} 2t+1 &= 3t\\ 4t^2 &=3t \end{align*}$$ have any common solutions. If they do, the two particles collide at that time. If they don't, the two particles don't collide (either their paths don't intersect, or they go through any points of intersection at different times).

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Observe all the motion from the frame of reference of any one particle say, the second particle ,

when you have changed your frame of reference , the data changes from this,

$x_1 (t) = 2t +1$ and $y_1 (t) = 4t^2$

$x_2(t) = 3t$ and $y_2 (t) = 3t$

to this,

$x_1 (t) = -t +1$ and $y_1 (t) = 4t^2-3t$

$x_2(t) = 0$ and $y_2 (t) = 0$

Now the question simply reduces to , whether particle 1 passes through origin or not?

equation of particle 1 is that of a projectile, i hope you can follow the rest from here.

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