The answer is: There will take 400 seconds to Sophia to lap Tyler.
To solve this problem we need to write equations in function to Tyler's distance when Sophia laps him, so:
Let be "x" the position of Tyler when Sophia is 400 m ahead of him.
Let be "x+400" the position of Sophia when she laps Tyler
Also, we must remember that time is equal to:
[tex]t=\frac{Distance}{V}[/tex]
So, the equation to calculate the position of the Tyler when Shopia laps him, can be written like this:
[tex]\frac{Tyler'sDistance}{Tyler'sSpeed}=\frac{Tyler'sDistance+400}{Sophia'sSpeed}[/tex]
Then,
[tex]\frac{x}{4\frac{m}{s} }=\frac{x+400m}{5\frac{m}{s}}\\\\(5\frac{m}{s})*(x)=(4\frac{m}{s})(x+400m)\\\\(5\frac{m}{s})*(x)=(4\frac{m}{s})*(x)+(4\frac{m}{s})*400m\\\\(5\frac{m}{s})*(x)-(4\frac{m}{s})*(x)=1600\frac{m^{2} }{s}\\\\(1\frac{m}{s})*(x)=1600\frac{m^{2} }{s}\\\\x=(1600\frac{m^{2} }{s})*\frac{s}{m}=1600m[/tex]
Therefore, the distance that Tyler's ridden when Sophia laps him is 1600m.
Let's calculate the time with the following formula:
[tex]Tyler'sDistance=x=xo+v*t\\\\1600m=0+4\frac{m}{s}*t\\\\t=\frac{1600m}{4\frac{m}{s} }=400s[/tex]
If we want to prove that the result is ok, let's substitutite the same distance in the Sophia's distance equation:
[tex]x+400m=xo+v*t\\\\1600m+400m=0+5\frac{m}{s} *t\\2000m=5\frac{m}{s} *t\\\\5\frac{m}{s} *t=2000m\\\\t=\frac{2000m}{5\frac{m}{s} }=400s[/tex]
Hence,
There will take 400 seconds to Sophia to lap Tyler.
Have a nice day!
