Asked by srikala
if a car covers 2/5th of the total distance with v1 speed and 3/5th distance with v2 then average speed is?
Answers
Answered by
Elena
The time
for the first part is
t1 =2•s/5•v1,
for the second part is
t2 = 3•s/5•v2.
The total time is
t =t1+t2 =
=(2•s/5•v1)+( 3•s/5•v2)=
=(s/5) •(2/v1 + 3/v2) =
=(s/5) •{(2•v2+3•v1)/(v1•v2)},
v(ave) = distance/total time =
=s/[(s/5) •{(2•v2+3•v1)/(v1•v2)}]=
=5•(v1•v2)/(3•v1+2v2).
for the first part is
t1 =2•s/5•v1,
for the second part is
t2 = 3•s/5•v2.
The total time is
t =t1+t2 =
=(2•s/5•v1)+( 3•s/5•v2)=
=(s/5) •(2/v1 + 3/v2) =
=(s/5) •{(2•v2+3•v1)/(v1•v2)},
v(ave) = distance/total time =
=s/[(s/5) •{(2•v2+3•v1)/(v1•v2)}]=
=5•(v1•v2)/(3•v1+2v2).
Answered by
sree sai
5v1v2/3v1+2*3/5
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