Question 159504
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Remember: {{{Speed=S=distance/time}}} ---> {{{time=distance/Speed}}}... EQN 1
{{{S[1]}}}=Going destination
{{{S[2]}}}=Returning back={{{highlight(2S[1])}}}, twice as fast as going --> Condition 1
IMPORTANT: {{{d[1]=90miles}}}, This distance is only for GOING for the trip.
ALSO: {{{t[1]+t[2]=3hours}}}, Total time going + time returning = 3 hours, in this eqn we get ----->{{{t[1]=3-t[2]}}}, EQN 2
IMPORTANT: See EQN 1, time and Speed are inversely proportional. What does it mean? When Speed goes up, time goes down. Let's illustrate with this proportion:
{{{S[1]/S[2]=t[2]/t[1]}}}
You see, when you double {{{S[2]}}}, then {{{t[1]}}} doubles and vice versa, that's why INVERSELY PROPORTIONAL (not direct, not {{{S[2]}}} then {{{t[2]}}})
Hope you got it because it's important when we apply it in getting {{{t[2]}}}
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With that said, as {{{S[2]=2S[1]}}}, then {{{t[1]=2t[2]}}},EQN 3, oks?
Then we can equate EQN 2 and EQN 3 since both equal to {{{t[1]}}}:
{{{3-t[2]=2t[2]}}}
{{{3=2t[2]+t[2]}}}
{{{3=3t[2]}}} -----> {{{cross(3)1/cross(3)1=cross(3)t[2]/cross(3)}}}
{{{t[2]=1hour}}}
And, via EQN 2: {{{t[1]=3-1=2hrs}}}
Therefore we can continue in getting {{{S[1]}}} as per EQN 1:
{{{S[1]=d[1]/t[1]=90miles/2hours=cross(90)45/cross(2)}}}
{{{S[1]=45miles/hr}}} -----------------------> Speed going for the trip
And, as per Condition 1:
{{{S[2]=2*45=90miles/hr}}} ------------------> Speed returning back
In doubt? go back  EQN 1:
{{{TotalTime=d[1]/S[1]+d[2]/S[2]}}}
{{{3hours=90/45+90/90}}}
{{{3hours=2hours+1hour}}}
{{{3hours=3hours}}}
Thank you,
Jojo</pre>