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# Crest Speed, Average Speed & Schedule Speed of an Electric Train

## Crest Speed of Electric Train

The maximum speed attained by an electric train during running is termed as its "crest speed". It is denoted by V_{m}.

## Average Speed of Electric Train

Average speed is defined as the mean of the speeds attained by the electric train from start to stop. In other words, average speed is defined as the ratio of the distance covered by the train between two stops to the actual time of run. The average speed of train is denoted by V_{a} and is given by,

$$\mathrm{\mathrm{Average\: speed,}V_{a}\:=\:\frac{\mathrm{Distance\: between \:two\: stops\: (\mathit{D})}}{\mathrm{Actual\: time\: of\: run (\mathit{T_{\mathrm{run}}})}}\:\mathrm{kmph}}$$

## Schedule Speed of Electric Train

The schedule speed of an electric train is defined as the ratio of the distance between two stops to the actual time of the run including the time for stop. The schedule speed of a train is denoted by *V _{s}* and is given by,

$$\mathrm{\mathrm{Schedule\: speed,}V_{S}\:=\:\frac{\mathrm{Distance\: between \:two\: stops\: (\mathit{D})}}{\mathrm{Actual\: time\: of\: run \:+\:Time\: for\: stop}}}$$

$$\mathrm{\therefore V_{S}\:=\:\frac{\mathit{D}}{\mathit{T_{\mathrm{run}}}\:+\:\mathit{T_{\mathrm{stop}}}}\:=\:\frac{\mathit{D}}{\mathit{T_{S}}}}$$

Where,*T _{s}* is the

**schedule time**.

## Numerical Example

An electric train has maximum speed of 80 kmph and the distance between the two stops is 6 km. Determine the average speed and schedule speed over the run assuming −

Free running of 300 seconds.

Duration of stops as 60 seconds.

**Solution**

Given,

Maximum speed,V

_{m}= 80 kmphDistance between two stops,

*D*= 6 kmActaul run time,

*T*_{run}= 300 seconds = $\frac{300}{3600}$hoursStop time,

*T*_{stop}= 60 seconds = $\frac{60}{3600}$ hours

Therefore, the **average speed** of the train is,

$$\mathrm{\mathrm{Average\: speed}\:V_{\mathit{a}}\:=\:\frac{\mathit{D}}{\mathit{T_{\mathit{run}}}}\:=\:\frac{6}{\mathrm{\left ( 360/3600 \right )}}\:=\:72\:\mathrm{kmph}}$$

Now, the schedule time is given by,

$$\mathrm{\mathit{T_{\mathit{S}}}\:=\:\mathit{T_{\mathrm{run}}}\:=\:\mathit{T_{\mathrm{stop}}}\:=\:300\:+60\:=\:360\:\mathrm{sec}\:=\:\frac{360}{3600}\:\mathrm{hours}}$$

Therefore, the **schedule speed** of the train is given by,

$$\mathrm{\mathrm{Schedule\: speed}\:V_{\mathit{S}}\:=\:\frac{\mathit{D}}{\mathit{T_{\mathit{S}}}}\:=\:\frac{6}{\mathrm{\left ( 360/3600 \right )}}\:=\:60\:\mathrm{kmph}}$$

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