Does this involve calculus? im confused help please

  • Thread starter pringless
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Hint: x_1 and x_2 never have the same value. The x-coordinates of two objects moving along the x-axis are given below as a function of time t. x_1 = (4m/s)t x_2 = -(25m) + (8m/s)t - (2m/s^2)t^2 Calculate the magnitude of the distance of closest approach of the two objects.
 

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  • #2
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Hi pringless,
no, you do not need calculus to solve this.
Let's call x(t) = x1(t) - x2(t).
This is just quadratic in t (the graph is a parabola).
All you got to do is find the lowest (or highest) point of the parabola.
Let's call that point (t0, x0), then we can write
x(t) = a(t - t0)2 + x0.
You can find a, t0, x0 by matching the coefficients on both sides. OK?
 
  • #3
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im sorry...i dont really understand what u mean
 
  • #4
enigma
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Pringless,

Graph the two equations. That will show you position of each particle as a function of time.

If you then subtract one from the other, you'll have the difference between the two. If you graph that, you'll see the difference as a function of time. You'll see that it will go down and then go back up. The closest approach is where the difference is the smallest.
 

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