Mathematics

Mathematics

Problem

A particle moves along a line with velocity v(t)=−3t2+12t−7v(t)=-3t^{2}+12t-7 meters per second for 0≤t≤40\le t\le4. Given s(0)=0s(0)=0, at what time tt does the particle attain its absolute minimum position on this interval? a) t=2t=2 b) t=0t=0 c) t=2−153t=2-\dfrac{\sqrt{15}}{3} d) t=2+153t=2+\dfrac{\sqrt{15}}{3}
Intuitive explanation

The particle's position is found by integrating velocity, but the minimum position isn't necessarily where velocity is zero—you need to check where the particle actually is at critical times. Find when v(t)=0v(t) = 0 to locate turning points, integrate to get s(t)s(t), then evaluate position at those turning points and the endpoints to see which gives the smallest value. The key is that the particle might reverse direction, so you're tracking its actual location, not just when it stops.

Intuitive explanation

The particle's position is found by integrating velocity, but the minimum position isn't necessarily where velocity is zero—you need to check where the particle actually is at critical times. Find when v(t)=0v(t) = 0 to locate turning points, integrate to get s(t)s(t), then evaluate position at those turning points and the endpoints to see which gives the smallest value. The key is that the particle might reverse direction, so you're tracking its actual location, not just when it stops.

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