Answer:
[tex]His\ average\ speed\ to\ trip\ there\ =45\ mph\\\\His\ average\ speed\ in\ return\ trip=45-9=36\ mph\\\\His\ average\ speed\ during\ whole\ trip=40\ mph[/tex]
Step-by-step explanation:
[tex]Let\ distance\ from\ one\ side=d[/tex]
Trip:
[tex]Let\ speed\ when\ he\ was\ going=x\ mph\\\\Time\ taken=4\ hours\\\\distance=d\\\\distance=speed\times time\\\\d=x\times 4\\\\d=4x\ ...................................eq(1)[/tex]
Return Trip:
[tex]Speed=x-9\ \ \ \ \ (as\ it\ is\ 9\ mph\ slower)\\\\Time\ taken=5\ hours\\\\Distance=d\\\\Distance=speed\times time\\\\Distance=5\times (x-9)\\\\d=5x-45\ ........................................eq(2)[/tex]
[tex]from\ eq(1)\ and\ eq(2)\\\\5x-45=4x\\\\5x-4x=45\\\\x=45\ mph\\\\from\ eq(1)\\\\d=4\times 45=180\ miles[/tex]
[tex]Average\ speed\ during\ whole\ trip=\frac{total\ distance}{total\ time}=\frac{2\times 180}{5+4}=\frac{360}{9}=40\ mph[/tex]
[tex]His\ average\ speed\ to\ trip\ there\ =45\ mph\\\\His\ average\ speed\ in\ return\ trip=45-9=36\ mph\\\\His\ average\ speed\ during\ whole\ trip=40\ mph[/tex]
