Given,
The mass of the train, m=1.5×10 kg=15 kg
The force with which the locomotive pulls the train, F=7.5×10⁵ N
The initial speed of the train, u=0 m/s
The final speed of the train, v=85 km/hr
The final speed in m/s is
[tex]\begin{gathered} v=\frac{85\times1000}{3600} \\ =23.6\text{ m/s} \end{gathered}[/tex]From Newton's second law,
[tex]F=ma[/tex]Where a is the acceleration of the train.
On substituting the known values,
[tex]\begin{gathered} 7.5\times10^5=15\times a \\ \Rightarrow a=\frac{7.5\times10^5}{15} \\ =50\times10^3m/s^2 \end{gathered}[/tex]From the equation of motion,
[tex]v=u+at[/tex]Where t is the time interval in which the train will accelerate to the given speed.
On substituting the known values,
[tex]\begin{gathered} 23.6=0+50\times10^3\times t \\ \Rightarrow t=\frac{23.6}{50\times10^3} \\ =4.72\times10^{-4}\text{ s} \end{gathered}[/tex]Thus the train will accelerate to the given speed in 4.72×10⁻⁴ s
