Mathematics

Mathematics

Problem

## End of Chapter 10 Drill The answers are in Chapter 24. 5. If the position function of a particle is x(t)=4t3−12t2+18,t>0x(t)=4 t^{3}-12 t^{2}+18, t>0, at what time is the particle changing direction? (A) t=0t=0 (B) t=12\quad t=\frac{1}{2} (C) t=1t=1 (D) t=2t=2 (A) t=0t=0 (B) t=12\quad t=\frac{1}{2} (C) t=1t=1 (D) t=2t=2
Intuitive explanation

A particle changes direction when its velocity switches sign—that is, when v(t)=0v(t) = 0 and the velocity actually reverses rather than just touching zero. Find the velocity by differentiating the position function, then solve v(t)=0v(t) = 0. The key is recognizing that a direction change happens at a critical point where the velocity crosses zero (not just touches it), which you can verify by checking the sign of velocity just before and after.

Intuitive explanation

A particle changes direction when its velocity switches sign—that is, when v(t)=0v(t) = 0 and the velocity actually reverses rather than just touching zero. Find the velocity by differentiating the position function, then solve v(t)=0v(t) = 0. The key is recognizing that a direction change happens at a critical point where the velocity crosses zero (not just touches it), which you can verify by checking the sign of velocity just before and after.

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