Question

The velocity of a car for a part of its journey is given by the function v(t) = -t^3 + 12t^2 - 20t + 10, where tea is the time and seconds when does the car attain it’s maximum velocity

Answer 1

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.

`\begin{gathered} v(t)=-t^3+12t^2-20t+10 \\ v(t)´=-3t^2+24t-20 \end{gathered}`

Now, let's find critical points, where the first derivative is equal to zero. We can use quadratic equation:

`\begin{gathered} -3t^2+24t-20=0 \\ a=-3\text{ }b=24\text{ }c=-20 \\ t=(-b\pm√(b^2-4ac))/(2a) \\ t=(-24\pm√(336))/(-6) \\ t_1=0.95 \\ t_2=7.05 \end{gathered}`

Now, let's find second derivative:

`v(t)^(\prime)^(\prime)=-6t+24`

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.

`\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}`

As you can see, when we use t=7.05 the result is negative. So, t=7.05 is a maximum of the function.