$v(t)=(3t^2-10t+4)e^{-0.5t}-4\cos t$, where $0\leq t\leq6$. Find the maximum displacement of the

Question

$v(t)=(3t^2-10t+4)e^{-0.5t}-4\cos t$, where $0\leq t\leq6$. Find the maximum displacement of the particle. Determine the intervals of time when the acceleration of the particle is negative.

Accepted Answer

(a) $4.81$ (b) $0\leq t<0.792$ and $3.50<t\leq6$

IB AA HL May 2026 P2 TZC Q5

IB AA HL May 2026 P2 TZC Q5

Solution:

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