Question

a person while.driving to office compute the average speed for his trip to be 70 km/h on his return trip along the same route there is less traffic and average speed is 50km/h what is average speed of persons trip

Answer 1

Answer:

Given :

Method I:  

Let the distance traveled by Abdul from home to school = s km

time taken to reach the school=t1 sec

For Return journey , Abdul cover distance =s km

time=t2 sec

∴ Average speed for forward journey[home - school ] = Total distance/ Total time.

20 km/h =s/t1

∴t1=s/20 h ------eq(1)

 Average speed for backward journey[school -home] = Total distance/ Total time.

30 km/h =s/t2

t2=s/30h -----eq(2)

Average distance for entire journey = Total distance/ Total time

                                                        =(s+s)/[s/20 +s/30]

                                                        =2s/s[1/20+1/30]

                                                         =2x20x30/50

                                                          =24 km/hr

Shortcut method : If equal distance are covered with speeds v1 and v2 then average distance=2v1v2/v1+v2

                          = 2x20x30/50

                           =24km/hr

∴ The average speed for persons trip is 24km/h

Answer 2

Given:-

Speed for the first case=70km/h

Speed for the second case=50km/h

Let the,two speeds be V1 and V2 respectively

To find:-

Average speed of the person's trip

Solution:-

We can see here that the person travels same distance in both cases.

Thus,when distance travelled is same,then average speed is------

$= > \frac{2v1v2}{v1 + v2}$

$= > \frac{2 \times 70 \times 50}{70 + 50}$

=>58.33km/h

Thus,the average speed is 58.33km/h.