Question

In a 500m race "A" beats "B" by 5seconds or 100 m.They decided to run another race and this time "A" gives "B" a head start of 200m.If "A" speed is twice his previous speed and B speed is one half of his previous speed,how far from the start should the winning post be so that they finish at same time?

Answer 1

I dont know if this is right, i might be interpreting the problem wrong >.<
anywhos, im thinking since A beat B by 5 secs, which was 100 m, then B's velocity is 100/5 = 20 m/s
Then, we know he ran the 500m in 5 more secs than A did so:
\[D = r_b(t_a+5) \iff 500 = 20(t_a+5) \iff t_a = 20 s\]
which gives A a rate of:
\[500 = r_a(20) \iff r_a = 25 m/s\]
Still working out the rest.

Answer 2

So now when they are running the second time:
\[r_{a'} = 2r_a = 50 m/s, r_{b'} = \frac{1}{2}r_b = 10 m/s\]
we have the distance is going to be:
\[d = r_{a'}t\iff d = 50t, d = r_{b'}t+200 \iff d = 10t+200\]
Solving this system gives:
d = 250 m.

Answer 3

Ans : 500 m

Answer 4

wut? >.> lol let me look over junks again.

Answer 5

i dont see how that answer makes sense =/ in the first scenario, A has to be running 25 m/s, and B has to be running 20 m/s in order for A to win by 5 secs (500m dash).
So in the second race, A will be running at 50 m/s, and B will be running at 10 m/s
if the answer was 500 m, A can run that in 10 secs, B would take 30 secs (he only has to run 300 m)

Answer 6

In the answer 250m, it would take A 5 secs to run it, and B 5 secs (he only has to run 50 m)

Answer 7

I just see the answer from answer key it is given in the book as 500m .I m not able to solve it.

Answer 8

i must be interpreting the problem wrong then. It has to do something with how im calculating their speeds in the first race.