Answer:
[tex]t=2.53[/tex]
Step-by-step explanation:
[tex]\text{Solution:}\\\text{Displacement function, }s=t^4-4t^3-8t^2+25t\\\text{Now,}\\\text{Velocity function, }v=\dfrac{ds}{dt}=\dfrac{d(t^4-4t^3-8t^2+25t)}{dt}\\\\\text{or, }v=4t^3-4(3t^2)-8(2t)+25\\\text{or, }v=4t^3-12t^2-16t+25[/tex]
[tex]\text{Acceleration function, }a=\dfrac{dv}{dt}\\\\\text{or, }a=\dfrac{d(4t^3-12t^2-16t+25)}{dt}\\\\\text{or, }a=4(3t^2)-12(2t)-16\\\text{or, }a=12t^2-24t-16[/tex]
[tex]\text{Now we need to calculate }t\text{ when }a=0.\\\therefore\ a=0\\\text{or, }12t^2-24t-16=0\\\\\text{Using the quadratic formula,}\\t=\dfrac{-(-24)\pm\sqrt{(-24)^2-4(12)(-16)}}{2(12)}=\dfrac{24\pm\sqrt{576+768}}{24}=\dfrac{24\pm8\sqrt{21}}{24}\\\text{or, }t=\dfrac{3\pm\sqrt{21}}{3}[/tex]
[tex]\text{Given that }t\ge0,\\\\\therefore\ t=\dfrac{3+\sqrt{21}}{3}\approx2.53[/tex]
Formula used:
[tex]1.\ \dfrac{dx^n}{dx}=nx^{n-1}\\\\2.\ \text{If }ax^2+bx+c=0\text{ and }a\ne0,\ x=\dfrac{-b\pm\sqrt{b^2-4ac}{}}{2a}[/tex]
