What average speed must be maintained for the last two laps in order to win?

A runner must sprint at 20m/s for four laps to win. However, the runner averages only 15m/s for the first two laps. What average speed must be maintained for the last two laps in order to win?





Please give me tips on how to work out the answer,any help would be great!


|||Let's Say 1 lap equals X meters.





Then you are given that


4X= 20* t where t is in seconds.


thus t= X/5 seconds....





To win the runner must do better than this time or at least have this much time.





Now You are also given that


2X= 15 * t1 (coz 1 lap equals X m)


hence t1= 2X/15





Now two laps are left let's say they are covered with average speed v then


2X= v * t2 which would give t2= 2X/v





Now total time is t1+t2 which should be %26lt;= t (original time with 20m/s for all 4 laps)





2X/V + 2X/15 %26lt;= X/5 (cancel out X and solve the inequality for V)





If you solve this you'll get V %26gt;= something,





which is correct since if the runner has to win he must have some minimum average speed.





The minimum speed would be if you remove the %26gt; sign and take only the equality sign i.e V= something (would be minimum speed)





Hope you understood this.|||We know that average speed is equal to the following:





avg speed = (total Distance)/(total Time)





So for the first two laps, he averages 15m/s


For the last two laps, he averages x m/s


For the entire four laps, he averages 20m/s





Let's say that each lap is a distance: d


Let's say that he completes the first two laps in t1 seconds


Let's say that he comletes the last two laps in t2 seconds





Now we have information to solve for the average speed for the last two laps:





15 = (2d)/(t1) 15 is the average speed of the first two laps (hence 2d ad the t1)





2d = 15t1


d = 15t1/2





20 = (4d)/(t1+t2)


20t1 + 20t2 = 4d we know that d = 15t1/2


20t1 + 20t2 = 4(15t1/2)


20t1 + 20t2 = 30t1


20t2 = 10t1


t2 = t1/2 meaning he must complete the last two laps in half the time it takes to complete the first two laps





Now we can find average speed for last two laps





avg speed = (2d)/t2 we know d and t2


avg speed = (2(15t1/2))/(t1/2)


avg speed = 15t1/(t1/2)


avg speed = 15t1 * (2/t1) t1's cancel out


avg speed = 15 * 2 = 30





voila