Running Back And Forth

The time taken by the platoon and the the last person to cover distances is the same. Hence their distances covered can be equated.

Lets say when the the last person reaches the front rank, the distance covered by the platoon is d.

so the last person will run (50+d) while the platoon moves d.

Now when he runs back to reach his original position, he will cover {50-(50-d)} =d while the platoon will move (50-d).

Thus equating,

(50+d)/d= d/(50-d)

d^2= 2500--d^2

d= 35.355

Distance covered by last person is (50+d)+d = 120.71

let speed of the last person be v1 wrt ground and that of platoon be v2 wrt ground. using relative velocities, { 50/(v1-v2) + 50/(v1+v2) } =50. this gives v1=v2(\sqrt{2}+1) So the final distance comes out as 50(1+\sqrt{2})=120.7

It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backword - are equal.

Let's assume that when the last person reached the first person, the platoon moved X meters forward.

Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.

Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon moved (50-X) meters.

Now, as the ratios are equal, (50+X)/X = X/(50-X) (50+X) (50-X) = X X

Solving, X=35.355 meters

Thus, total distance covered by the last person = (50+X) + X = 2 X + 50 = 2 (35.355) + 50 = 120.71 meters

Note that at first glance, one might think that the total distance covered by the last person is 100 meters, as he ran the total lenght of the platoon (50 meters) twice. TRUE, but that's the relative distance covered by the last person i.e. assuming that the platoon is stationary.

According to the question,

Time required for the Last man to run 'along' the parade and reach the 1st person + Time required for the Last man to run 'against' the parade and reach his original position = Time required by the parade to cover 50 m

let 'x' be the running speed of the last person and y be the parade speed

This gives, 50/(x-y) + 50(x+y) = 50y (here 50 m is the length of the parade, not the distance covered by the parade)

,we get (x/y)^2 - (x/y) - 1 = 0.....hence x/y = (2.414/1)

From the ratio of x/y, if 1 m/s is the parade speed, it will take takes 50 seconds to cover 50 m. So, the last person runs a distance of 2.414*50 = 120.7 m :)

Let platoon speed be v and person's speed be u.Time for whole of the process is 50/v. So we need to find 50 (u/v) According to the condition [50/(u-v)] +[50/(u+v)]=50/v which gives 2uv=u^2 - v^2... Dividing the equation by v^2 we get quadratic in u/v solving which we get u/v=1+root 2. Now you can find 50 (u/v)..