If a car covers 2/5th of the total distance with speed V¹ & 3/5th distance with speed V², then what will be the average speed?
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Let the total distance be x.
Time taken in travelling 2x/5 = 2x/5v1.
Time taken in travelling 3x/5 =3x/5v2.
Total time taken = 2x/5v1 + 3x/5v2 =(2xv2 +3xv1)/5v1v2.
Average speed = Total distance/Total Time
Therefore Av. speed = (x × 5v1v2)/(2xv2 + 3xv1)
=5v1v2/(3v1 + 2v2).
Time taken in travelling 2x/5 = 2x/5v1.
Time taken in travelling 3x/5 =3x/5v2.
Total time taken = 2x/5v1 + 3x/5v2 =(2xv2 +3xv1)/5v1v2.
Average speed = Total distance/Total Time
Therefore Av. speed = (x × 5v1v2)/(2xv2 + 3xv1)
=5v1v2/(3v1 + 2v2).
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