Question

an electric train is to have a braking retardation of 3.2 kmph. the ratio of maximum speed is to average is 1.3 , the time for stop is 26 seconds and acceleration is 0.8 kmph. the run is 1.5 km. actual time of run is​

Answer 1

Answer:

Step-by-step explanation:

Given: retardation = 3.2 km/h (this is speed just before stopping)

         acceleration = 0.8 km/h (this is speed just after starting)

Ratio of maximum speed to average = 1.3

Time for stop = 26 seconds ( in this time the speed comes from 3.2 km/h to 0 km/h)

Distance of the run = 1.5 km

To find: actual time of run

we have, time = $\frac{ distance}{Average speed}$  ⇒ 1

Now, we have, maximum speed : average speed = 1.3

i.e, $\frac{max speed}{avg speed}$ = 1.3

∴ avg speed = $\frac{max speed}{1.3}$

From acceleration and retardation, we find that the speed of retardation

is more. Hence that is the maximum speed.

∴ avg speed = $\frac{3.2}{1.3}$ =  $\frac{3.2 X 10}{1.3 X 10}$ = $\frac{32}{13}$

Inserting this value in equation 1,

time = $\frac{ distance}{Average speed}$  = $\frac{1.5}{\frac{32}{13} }$ = $\frac{15}{10}$×$\frac{13}{32}$ = $\frac{195}{320}$ = 0.603 hours

Converting into minutes =  $\frac{195}{320}$  x 60 = 18.87 minutes