An 18-year-old runner can complete a 10.0-km coursewith an average speed of 4.44

Question

An 18-year-old runner can complete a 10.0-km coursewith an average speed of 4.44 m/s. A 50-year-old runner can coverthe same distance with an average speed of 4.03 m/s. How much later(in seconds) should the younger runner start in order to finish thecourse at the same time as the olderrunner?
I thought this was going to be an easy problem but got itwrong. What I did was convert 10km into meters which is 10,000m.Then plugged the rest of the numbers into the average speedequation and solved for (t) to get the seconds for bothrunners finishing the course and then subtracted the 50 year oldstime from the 18 year olds time to get the seconds that the 18 yearold must start after the 50 year old starts running. Which was 209seconds, can someone tell me what went wrong? An 18-year-old runner can complete a 10.0-km coursewith an average speed of 4.44 m/s. A 50-year-old runner can coverthe same distance with an average speed of 4.03 m/s. How much later(in seconds) should the younger runner start in order to finish thecourse at the same time as the olderrunner?
I thought this was going to be an easy problem but got itwrong. What I did was convert 10km into meters which is 10,000m.Then plugged the rest of the numbers into the average speedequation and solved for (t) to get the seconds for bothrunners finishing the course and then subtracted the 50 year oldstime from the 18 year olds time to get the seconds that the 18 yearold must start after the 50 year old starts running. Which was 209seconds, can someone tell me what went wrong?

Explanation / Answer

   Given that the total distance is S = 10.0km =10*103 m    The speed of the 18-year-old runner is V18 =4.4m/s    The speed of the 50-year-old runner is V50 =4.03m/s --------------------------------------------------------------------     Time taken by 18-year-old to travel thedistance S is t1 = S / V18                                                                                          = 10*103m / 4.4m/s                                                                                           =2272.72 s     Time taken by 50-year-old to travel thedistance S is t2 = S / V50                                                                                            = 10*103m / 4.03m/s                                                                                           = 2481.38 s     Then the difference in time is t = t2 -t1                                                   =208.66s                                                                                          = 10*103m / 4.03m/s                                                                                           = 2481.38 s     Then the difference in time is t = t2 -t1                                                   =208.66s     

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.