Question

Albert left NY on his bike riding at an average rate of 10 mph three hours before his son left by automobile. The son overtook Albert in exactly 30 minutes. At what average rate was Albert's son travelling?

Answer 1

Average rate Albert's son was traveling thus happens to be
`70mph`

Explanation:

Let the velocity of albert be #v1 #
Let the velocity of his son be #v2 #
Journey time of albert be #t1 #
Journey time of his son be #t2 #
Distance traveled by albert be #s1 #
Distance traveled by his son be #s2 #

Relations
`s1`=`v1.t1`
`s2`=`v2.t2`
When the son overtook albert,
`s1`=`s2`
Equating the distances
`v1.t1`=`v2.t2`
Knowing that
`v1`=`10 mph`
and
Father leaves earlier than son by `3` hours,
`t1-t2`=`3`
Additional time for albert is
`t1`=`t2+3`
Equating the distances travelledby albert and his son,
`10.t1`=`v2.t2`
`t2`=`30 min`
`30min`=`0.5 hours`
`t2`=`0.5 hours`
`t1`=`0.5+3`
=`3.5 hours`
Distance traveled by albert is
`s1`=`10` x `3.5`
=`35` miles
Time taken by his son is `0.5 hour`
Average rate Albert's son was traveling thus happens to be
`35/0.5` = `70mph`