Question

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. The maximum speed is to be limited to 64 kmph, acceleration is 2 kmph and coasting is 0.16 kmph and braking retardations is 3.2 kmph The duration of costing is​

Answer 1

Step-by-step explanation:

sorry idont know sorry I don't know

Answer 2

Answer:

$97s.$

Step-by-step explanation:

Given, Distance between two stations $=1.6km$

Average speed $=43kmph$

Maximum speed $=64kmph$

Acceleration $=2kmph$

Coasting retardation $=0.16kmph$

Braking retardation $=3.2kmph$

Now,

The duration of the acceleration $=\frac{V_m}{a} =\frac{64}{2} =32s$

The actual time of run,  $T=\frac{3600*D}{V_a}$

                                      $T=\frac{3600*1.6}{43} =144s$

Before applying brakes; let the speed be $V_2$

The duraiton of coasting $=\frac{V_m-V_2}{B_c}$

                                     $t_2=\frac{64-V_2}{0.16} s$

The duration of braking $t_3=\frac{V_2}{B} =\frac{V_2}{3.2} s$

The actual time of run, $T=t_1+t_2+t_3$

                                 $144=32+\frac{64-V_2}{0.16}+\frac{V_2}{3.2}$

                                 $144=400-6.25V_2+0.313V_2$

                            $5.937V_2=288$

                                    $V_2=\frac{288}{5.937}=48.48kmph$

The duration of coasting,

                                    $t_2=\frac{64-48.48}{0.16} =97s.$