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A particle moves west at a speed of $25km/hr$ from the origin when $t=0$. Another particle moves north at a speed $20km/hr$ and stops at the origin when $t=1$, where $t$ is in hours. What will the minimum distance be between the $2$ particles and at what time $t$ does such minimum occur.

I have studied these questions for a while but i need help deriving an equation relating the two particle's movements to minimize them. Thanks for any help.

Note: The particles are co-planar.

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  • $\begingroup$ Write down parametric equations for each particle's position as a function of $t$. Subtract them. Use the usual Pythagoreal formula to get their distance as a function of $t$. Note that it will be somewhat easier to minimize the square of their distance than the distance itself. $\endgroup$
    – hmakholm left over Monica
    Nov 12, 2015 at 10:47
  • $\begingroup$ so far i have gotten this: $x_1(t)=-25t$ and $x_2(t)=20t-20$ $\endgroup$
    – Brayden
    Nov 12, 2015 at 10:50

2 Answers 2

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The particles can be viewed as one moving along the negative x-axis away from the origin, the other moving towards the origin along the negative x-axis. The equations for position can be given as $ r_1=(-25t,0)$ and $r_2=(0, 20t-20)$

This distance between them is the hypoteneuse of the right angles triangle formed by the x coordinates of the first particle and the y coordinate of the second. This means the distance between them is $$ \left( 25^2 t^2 + (20t-20)^2 \right) $$ If you find the minimum value of this for $0 \leq t \leq 1 $, you have your answer.

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Use a right triangle to visualize this situation. You know certain facts 1) the speed of the particles can be represented on the two legs 2) the hypotenuse can be the distance between the two particles

Thus, the function representing the distance will be based on the Pythagorean theorem.

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  • $\begingroup$ Ok i have gotten this so far, $d^2 = 25t^2+(20t-20)^2$, which expands to the quadratic $425t^2-800t+400$ $\endgroup$
    – Brayden
    Nov 12, 2015 at 10:53
  • $\begingroup$ which is wrong. $\endgroup$
    – Brayden
    Nov 12, 2015 at 10:56

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