Question

A body covers the first half of the distance
between two places at 40 m/s and second half at 60m/s
what is the average Speed of
the body​

Answer 1

Explanation:

Avg speed=

total distance covered/ total time taken

Suppose the total distance=s

Time taken for first half t1=distance/speed=(s/2)/v1

=S/2v1

Time taken for second half=s/2v2

Total time=s/2v1)+(s/2v2)=s/2(1/v1+1/v2)

=S(v1+v2)/2v1v2

Avg speed= S/ total time taken

By solving this you get

Avg speed=2v1v2/(v1+v2)

So for this question=

2″40″60/100=48m/s

Short trick

Assume the distance as lcm of 40 and 60

240 total distance=240

Time taken in first half=120/40=3

Second half=120/60=2

Avg speed=total distance/total time taken

=240/5=48m.

Answer 2

$\huge{\pink{\underline{\underline{Explanation-}}}}$

Average speed

Total distance covered / total time taken

suppose distance is equal to ‘s’.

time taken for the first half t1 =

distance /speed = (s/2)/v1

s=2v1

time taken for second half=s/2v2

total time = ( s/2v1) + ( s/2v2)

= s/2 (1/v1+1/v2)

= s(v1+v2)/2v1v2

Average speed = s/ total time taken

By solving this u gate,

average speed=2v1v2(v1v2)

so, for this question=

2"40" 60/10048m/s

SHORT TRICK

Assume the distance as LCM of 40 and 60

Total distance is 240.

time taken in first half= 120/40=3

time taken in second half= 120/60=2

average speed is = Total distance / total time taken.

= 240/5

=48m

HOPE IT HELPS.............✌️