$52.$ A train runs at an average speed of $50$ kmph between stations situated $2\mathrm{.}5$ km $,$ apart. Train accelerates at $2$ kmphps and retards at $3$ kmphps $.$ Find its maximum speed assuming simplified trapezoidal speed time curve. Draw the speed time curve for the run and calculate also the distance travelled by it before the brakes are applied. $($Ans$.\backslash (57\mathrm{.}75\backslash$mathrm$\{$kmph$\},2\mathrm{.}35\backslash$mathrm$\{$~km$\}\backslash ))$

$53.$ An electric train is to have a braking retardation of $3\mathrm{.}2$ kmphps $.$ If the ratio of maximum speed to average speed is $1\mathrm{.}3,$ the time for stops is $26$ seconds and acceleration is $0\mathrm{.}8$ kmphps $,$ find the schedule speed for a run of $1\mathrm{.}5$ km $.$ Assume simplified trapezoidal speed time curve. $($Ans$.30$ kmph $)$

$54.$ A train is required to run between two stations $16$ km apart at an average speed of $43$ kmph $.$ The run is to be made to a simplified quadrilateral speed time curve. If the maximum speed is to be limited to $64$ kmph $,$ acceleration to $2$ kmphps and coasting and braking retardations to $0\mathrm{.}16$ and $3\mathrm{.}2$ kmphps respectively, determine the duration of acceleration, coasting and braking periods. $($Ans$.\backslash (\backslash$mathrm$\{$t$\}\_\{1\}=32\backslash$mathrm$\{$sec$\},\backslash$mathrm$\{$t$\}\_\{2\}=96\mathrm{.}8\backslash$mathrm$\{$sec$\},\backslash$mathrm$\{$t$\}\_\{3\}\backslash )\backslash (=15\mathrm{.}15\backslash$mathrm$\{$sec$\}\backslash ).$

$55.$ A $250$ ton train with $\backslash (10\backslash \%\backslash )$ rotational inertia effect is started with uniform acceleration and reaches a speed of $50$ kmph in $25$ seconds on level rail road. Find the specific energy consumption if the journey is to be made according to simplified trapezoidal speed time curve, the acceleration is $2$ kmphps $,$ braking retardation $3$ kmphps and distance between two stations is $2\mathrm{.}4$ km $,$ efficiency of motors is $0\mathrm{.}9,$ track resistance is $\backslash (5\backslash$mathrm$\{$~kg$\}/\backslash$mathrm$\{$ton$\}\backslash ).$

$($Ans$.28$ Whr$/$ton$-$km$)$