Solution. B) 7.05 seconds
Analysis: We can use second derivative to find value of t where we would have maximum velocity.
In the first step, let's find first derivative.
[tex]\begin{gathered} v(t)=-t^3+12t^2-20t+10 \\ v(t)´=-3t^2+24t-20 \end{gathered}[/tex]Now, let's find critical points, where the first derivative is equal to zero. We can use quadratic equation:
[tex]\begin{gathered} -3t^2+24t-20=0 \\ a=-3\text{ }b=24\text{ }c=-20 \\ t=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ t=\frac{-24\pm\sqrt{336}}{-6} \\ t_1=0.95 \\ t_2=7.05 \end{gathered}[/tex]Now, let's find second derivative:
[tex]v(t)^{\prime}^{\prime}=-6t+24[/tex]Now, let's replace both critical points in second derivative. If the result is negative, the point is a maximum of the function. If the result is positive, the point is a minimum of the function.
[tex]\begin{gathered} t_1=0.95 \\ -6t+24 \\ -6(0.95)+24 \\ v(t)^{\prime}^{\prime}=18.3 \\ \\ t_2=7.05 \\ -6t+24 \\ -6(7.05)+24 \\ v(t)^{\prime\prime}=-18.3 \end{gathered}[/tex]As you can see, when we use t=7.05 the result is negative. So, t=7.05 is a maximum of the function.