Distance Math Word Problem?
A man "A" sets out from a certain point and traveled at the rate of 6 kph. After "A" had gone two hours , another man "B" set out to overtake "A" and went 4 km the first hour, 5 km the second hour, 6 km the third hour and so gaining 1 km every hour. How many hours after...

answers:
Captain Matticus, LandPiratesInc say: A = 12 + 6t
B = 4 + 5 + 6 + 7 + ...
B = 4 + 0 + 4 + 1 + 4 + 2 + 4 + 3 + 4 + 4 + 4 + 5 + ... + 4 + t
B = 4 + 4t + 1 + 2 + 3 + 4 + 5 + .... + t
B = 4 + 4t + (t/2) * (t + 1)
A = B
12 + 6t = 4 + 4t + (t/2) * (t + 1)
8 + 2t = (t/2) * (t + 1)
16 + 4t = t * (t + 1)
16 + 4t = t^2 + t
0 = t^2 + t  4t  16
0 = t^2  3t  16
t = (3 +/ sqrt(9 + 64)) / 2
t = (3 +/ sqrt(73)) / 2
t > 0
t = (3 + sqrt(73)) / 2
(3 + sqrt(73)) / 2 hours after B left their starting point
12 + 6t =>
12 + (6/2) * (3 + sqrt(73)) =>
12 + 3 * 3 + 3 * sqrt(73) =>
21 + 3 * sqrt(73) km from the starting point

Ash say: Lets say B meet A after t hours
then distance travelled by A = 6(t+2) km ...(1)
B's distance = 4 + 5 + 6...(t+3) km
which can be written as (3+1) + (3+2) + (3+3)...(t+3) ... where 2nd number in the sum is hour
This is arithmetic series with a₁=4, a(n)= (t+3), d = 1, n = t and
sum = n(a₁+a(n))/2 = t(4+(t+3))/2 = (t²+7t)/2 ...(2)
Since both distances will be same
6(t+2) = (t²+7t)/2
12(t+2) = (t²+7t)
12t + 24 = t²+7t
t²5t24=0
(t8)(t+3)=0
t = 8 or t=3
Ignoring negative time, our answer is 8 hours
B will meet A after 8 hours
