Question 1077331
Let {{{ s }}} = the speed of the plane with no wind in km/hr
Let {{{ w }}} = the speed of the headwind in km/hr
{{{ 2w }}} = the speed of the tailwind
{{{ 20 }}} min = {{{ 1/3 }}} hrs
---------------------------------------
For Toronto to Vancouver:
(1) {{{ 3900 = ( s - w )*5 }}}
For Vancouver to Toronto:
(2) {{{ 3900 = ( s + 2w )*(13/3) }}}
--------------------------------
(1) {{{ 780 = s - w }}}
(1) {{{ s = 780 + w }}}
Plug this result into (2)
(2) {{{ 3900 = ( 780 + w + 2w )*(13/3) }}}
(2) {{{ 3900 = ( 780 + 3w )*(13/3) }}}
(2) {{{ 11700 = 10140 + 39w }}}
(2) {{{ 39w = 1560 }}}
(2) {{{ w = 40 }}}
and
{{{ 2w = 80 }}}
-----------------------
(a)
the headwind is 40 km/hr
The tailwind is 80 km/hr
------------------------
(b)
First I have to find {{{ s }}}
(1) {{{ 3900 = ( s - w )*5 }}}
(1) {{{ 3900 = ( s - 40 )*5 }}}
(1) {{{ 3900 = 5s - 200 }}}
(1) {{{ 5s = 4100 }}}
(1) {{{ s = 820 }}}
Let {{{ t }}} = time between the cities without wind
For Toronto to Vancouver:
(1) {{{ 3900 = s*t }}}
(1) {{{ 3900 = 820t }}}
(1) {{{ t = 4.756 }}} hrs
and
For Vancouver to Toronto:
(2) {{{ 3900 = s*t }}}
(2) {{{ 3900 = 820*t }}}
I get the same answer
--------------------------
check my math & get another opinion