## Average Speed

Average speed is always equal to total distance travelled to total time taken to travel that distance.Average speed=Total distanceTotal timeAverage speed=Total distanceTotal time

**Distance Constant**

If distance travelled for each part of the journey, ie d1=d2=d3=…=dn=dd1=d2=d3=…=dn=d, then**average speed of the object is Harmonic Mean of speeds**.

Let each distance be covered with speeds s1,s2,…sns1,s2,…sn in t1,t2,…tnt1,t2,…tn times respectively.

Then t1=ds1t1=ds1, t2=ds2t2=ds2, … tn=dsntn=dsn

Average Speed=d+d+d+…ntimes(ds1)+(ds2)+(ds3)+…(dsn)Average Speed=d+d+d+…ntimes(ds1)+(ds2)+(ds3)+…(dsn)Average Speed=n(1s1)+(1s2)+…+(1sn)

## Important shortcuts to solve time and distance problems quickly

Using the shortcuts provided below, you can solve the aptitude problems on time and distance quickly

- Given a person covers a distance with speed
**a km/hr**and further covers same distance with speed**b km/hr**, then the average speed of the person is:

Average speed=Total distance travelledTotal time takenAverage speed=Total distance travelledTotal time taken

Let the distance covered be**d km**.

Given**d km**be covered with speed**a km/hr**in time t1t1 hour => t1=dat1=da

Given next**d km**be covered with speed**b km/hr**in time t2t2 hour => t2=dbt2=db

Average speed=2dda+dbAverage speed=2dda+db

Average speed=2aba+bAverage speed=2aba+b

Shortcut: As discussed in “Basic Concepts” section, average speed is the HM (Harmonic Mean) of speeds a & b

Formula

Example

- A train is 200 long runs at 50 m/s than find the time?

2) A train of length 160 m is running at a speed of 48 km/h while running its front and reaches a pole , How much time this dose the train take to completely pass the pole