Question 638443

Two contestants run a 440-yard course. The first wins by 4 seconds when given a head start of 200 feet. They finish together when the first is given a head start of 40 yards. Find the speed of each in feet per second. 

I tried to solve by using the formula D (distance)= s (speed) x t (time). 
But I come up with two equations for each situation. Converting yards to feet: 440yards=1320 feet: 
runner #1: s(t-4)=1120   (wins by 4 seconds)
runner #2: s(t)=1320     

and for the second situation:
runner #1: s(t)=1200 (given a 40 yard=120feet head start:  1320-120=1200 feet). 
runner #2  s(t)=1320
then solving for s: s=1320/t and substituting 1320/t(t-4)=1120 and solving for t, 1320t-5280=1120t, t=26.4 secs, then s=1320/t and s=1320/26.4= 50ft/sec. 

Now, obviously a human cannot run at that speed, so this does not seem like a plausible answer, and I cannot figure out why I have four equations, I also tried to use different combinations but this didn't work either. So, I trust someone with better mathematical acumen can come up with a better solution. Thank you so kindly for trying to solve this problem. I look forward to the correct solution and appreciate your time and assistance. 


Let the speed of the 1st runner be {{{S[1]}}}, and speed of 2nd, {{{S[2]}}}
The course is 440 yards, or 1,320 feet
Since the 1st gets a headstart of 200 feet, then the 1st travels 1,320 – 200, or 1,120 feet


Time taken by 1st to go 1,120 feet, plus time won by, equals time taken by 2nd to go 1,320 ft, OR

{{{1120/S[1] + 4 = 1320/S[2]}}}
 
{{{1120S[2] + 4S[1]S[2] = 1320S[1]}}} ---- Multiplying by LCD, {{{S[1]S[2]}}} ---- eq (i)



Since the 1st gets a headstart of 40 yards, or 120 feet, then the 1st travels 1,320 – 120, or 1,200 feet


Time taken by 1st to go 1,200 feet equals time taken by 2nd to go 1,320 ft, OR

{{{1200/S[1] = 1320/S[2]}}}
 
{{{1200S[2] = 1320S[1]}}} ---- Cross-multiplying ---- eq (ii)


Therefore, {{{1120S[2] + 4S[1]S[2] = 1200S[2]}}} ---- ----- Substituting {{{1200S[2]}}} for {{{1320S[1]}}} in eq (i)
{{{4(280S[2] + S[1]S[2]) = 4(300S[2])}}} ----- Factoring out GCF, 4

{{{280S[2] + S[1]S[2] = 300S[2]}}}
 
{{{S[1]S[2] = 300S[2] - 280S[2]}}}
 
{{{S[1]S[2] = 20S[2]}}}

{{{S[2](S[1]) = S[2](20)}}} ---- Factoring out GCF, {{{S[2]}}}

{{{S[1]  = 20}}}


{{{S[1]}}}, or speed of 1st runner = {{{highlight_green(20)}}} ft/s 


{{{1200S[2] = 1320(20)}}} ---- Substituting 20 for {{{S[1]}}} in eq (ii)

{{{1200S[2] = 26400}}}

{{{S[2]}}}, or speed of 2nd runner = {{{26400/1200}}}, or {{{highlight_green(22)}}} ft/s


You can do the check!!


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