First, find the acceleration using Newton's Second Law.
[tex]\begin{gathered} F=ma \\ a=\frac{F}{m} \end{gathered}[/tex]Where F = 8.10x10^5 N and m = 1.40x10^7 kg.
[tex]a=\frac{8.10\times10^5N}{1.40\times10^7\operatorname{kg}}=5.79\times10^{-2}\cdot\frac{m}{s^2}[/tex]Then, use a formula that relates acceleration, initial velocity, final velocity, and time.
[tex]v_f=v_0+at[/tex]Solve for t because the problem is asking to find the time.
[tex]\begin{gathered} v_f-v_0=at_{} \\ t=\frac{v_f-v_0}{a} \end{gathered}[/tex]Where vf = 64 km/h, v0 = 0, and a = 5.79x10^-2 m/s^2. Before we continue, we need to transform the final velocity to m/s.
[tex]v_f=64\cdot\frac{km}{h}\cdot\frac{1000m}{1\operatorname{km}}\cdot\frac{1h}{3600\sec}=17.78\cdot\frac{m}{s}[/tex]Once we have the velocity transformed, we are able to find t.
[tex]\begin{gathered} t=\frac{17.78\cdot\frac{m}{s}-0}{5.79\times10^{-2}\cdot\frac{m}{s^2}} \\ t=3.07\times10^2\sec \\ t=307\sec \end{gathered}[/tex]But, the answer must be in minutes.
[tex]t=307\sec \cdot\frac{1\min}{60\sec}=5.12\min [/tex]Therefore, it takes 5.12 minutes.