## does this involve calculus? im confused help please

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.
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
 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?
 im sorry...i dont really understand what u mean

Recognitions:
Gold Member
Staff Emeritus

## does this involve calculus? im confused help please

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.