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
- 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