Question 1058008
In both these cases, you can think of
it as one of the cyclists moving and the
other one standing still.
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When they are moving towards eachother,
the moving one moves at the sum of their speeds
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When moving in the same direction, let {{{ d }}} =
the distance the slower one travels until the
faster one catches up.
Moving towards eachother:
(1) {{{ 25 = ( s[1] + s[2] )*( 50/60 ) }}}
Moving in the same direction:
Slower one:
(2) {{{ d = s[2]*5 }}}
Faster one:
(3) {{{ d + 25 = s[1]*5 }}}
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(1)  {{{ (6/5)*25 = s[1] + s[2] }}}
(1) {{{ s[1] + s[2] = 30 }}}
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(2) {{{ s[2] = (1/5)*d }}}
(3) {{{ s[1] = (1/5)*d + 5 }}}
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plug (2) and (3) into (1)
(1) {{{ (1/5)*d + 5 + (1/5)*d = 30 }}}
(1) {{{ d + 25 + d = 150 }}}
(1) {{{ 2d = 125 }}}
(1) {{{ d = 62.5 }}}
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(2) {{{ d = 5*s[2] }}}
(2) {{{ 62.5 = 5s[2] }}}
(2) {{{ s[2] = 12.5 }}}
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(3) {{{ s[1] = (1/5)*d + 5 }}}
(3) {{{ s[1] = (1/5)*62.5 + 5 }}}
(3) {{{ s[1] = 12.5 + 5 }}}
(3) {{{ s[1] = 17.5 }}}
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The faster one's speed is 17.5 km/hr
The slower one's speed is 12.5 km/hr
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check answer:
(1) {{{ (6/5)*25 = s[1] + s[2] }}}
(1) {{{ (6/5)*25 = 17.5 + 12.5 }}}
(1) {{{ 30 = 30 }}}
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(2) {{{ d = s[2]*5 }}}
(2) {{{ d = 12.5*5 }}}
(2) {{{ d = 62.5 }}} km
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(3) {{{ d + 25 = s[1]*5 }}}
(3) {{{ d + 25 = 17.5*5 }}}
(3) {{{ d = 87.5 - 25 }}}
(3) {{{ d = 62.5 }}} km
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OK