Question

In a race over a distance s, runner A starts from rest and accelerates at a1 for the first distance x and then runs at constant speed. Runner B starts from rest and accelerates at a2 for the first distance x and then runs at constant speed. Runner A begins running as soon as the race begins but B first takes a nap to rest up.What is the longest nap B can take and still not lose the race?

Answer 1

Answer:

    dt = (x+s)/sqrt(2x)*(1/sqrt(a1) - 1/sqrt(a2))

Explanation:

- Equation of motion for constant acceleration are:

                                        v = a*t

                                     z =( a*t^2)/2  

                                     v^2 = 2*z*s

- Compute time t1:

                                   x = (a*t^2)/2

                                    t_1 = sqrt(2*x/a)

- Velocity @ time t is: v = a*t = a*sqrt(2*x/a)

- Remaining distance (s-x) is run at above calculated speed, so the time t_2 is:

                             t_2 = (s-x)/v = (s-x)/(a*sqrt(2*x/a))

- Compute total T:

                    T = t_1 + t_2 = sqrt(2*x/a) + (s-x)/(a*sqrt(2*x/a))

- Simplify:                   T = (x+s)/sqrt(2*a*x)

                 

- Time taken by each runner:

                                 T_A =(x+s)/sqrt(2*a1*x)

                                T_B = (x+s)/sqrt(2*a2*x)+ dt

Equating the total times (T_A = T_B):  

                      (x+s)/sqrt(2*a2*x)+ dt = (x+s)/sqrt(2*a1*x)

Hence,

                   dt = (x+s)/sqrt(2*a1*x) - (x+s)/sqrt(2*a2*x))

                   dt = (x+s)/sqrt(2x)*(1/sqrt(a1) - 1/sqrt(a2))