Respuesta :
Answer:
Explanation:
Here is what we know;
Both Kathy and Cheryl cover the same distance ( one course).
Kathy completes the course in 3.6 hours.
Kathy's speed is 2 miles/hour faster than Cheryl's
Cheryl completes the course in 6 hours
Cheryl's speed is yet unkown.
Now, the speed v is defined as
[tex]v=\frac{D}{t}[/tex]where D is the distance covered and t is the time taken.
Now, let us say D = distance of one course. Then in Kathy's case, we have
[tex]v_{\text{kathy}}=\frac{D}{3.6hr}[/tex]since Kathy's speed is 2 miles per hour faster than Cheryl's, we have
[tex]v_{\text{kathy}}=\frac{D}{3.6}=2+v_{\text{cheryl}}[/tex]For Cheryl, we know that
[tex]v_{\text{cheryl}}=\frac{D}{6hr}[/tex]or simply
[tex]v_{\text{cheryl}}=\frac{D}{6}[/tex]Putting this into the equation for Kathy's speed gives
[tex]v_{\text{kathy}}=\frac{D}{3.6}=2+v_{\text{cheryl}}\Rightarrow v_{\text{kathy}}=\frac{D}{3.6}=2+\frac{D}{6}[/tex][tex]\Rightarrow\frac{D}{3.6}=2+\frac{D}{6}[/tex]We have to solve for D, the distance of a course.
Subtracting D/6 from both sides gives
[tex]\frac{D}{3.6}-\frac{D}{6}=2[/tex][tex](\frac{1}{3.6}-\frac{1}{6})D=2[/tex][tex]\frac{1}{9}D=2[/tex][tex]D=18\text{miles}[/tex]Hence, the distance of a course is 18 miles.
With the value of D in hand, we can now find the velocity of Kathy and Cheryl.
[tex]v_{\text{cheryl}}=\frac{D}{6hr}=\frac{18\text{miles}}{6hr}[/tex][tex]\Rightarrow\boxed{v_{\text{cheryl}}=3\text{ miles/hr}}[/tex]Hence, Cheryl's speed is 3 miles/hr.
Next, we find Kathy's speed.
[tex]v_{\text{kathy}}=\frac{D}{3.6hr}=\frac{18\text{miles}}{3.6hr}[/tex][tex]\boxed{v_{\text{kathy}}=5\text{miles}/hr\text{.}}[/tex]Hence, Kathy's speed is 5 miles/hr.
Therefore, to summerise,
Kathy's speed = 5 miles/hr
Cheryl's speed = 3 miles/hr
