SOLUTION: From the same place at 7am,A started walking in north at the speed of 5km/hr. After 1hour B started cycling in the east at the speed of 16km/hr. At what time will they will be at d

Algebra ->  Rate-of-work-word-problems -> SOLUTION: From the same place at 7am,A started walking in north at the speed of 5km/hr. After 1hour B started cycling in the east at the speed of 16km/hr. At what time will they will be at d      Log On


   

Question 1066541: From the same place at 7am,A started walking in north at the speed of 5km/hr. After 1hour B started cycling in the east at the speed of 16km/hr. At what time will they will be at distance of 52km apart from each other.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
From the same place at 7am,A started walking in north at the speed of 5km/hr.
After 1hour B started cycling in the east at the speed of 16km/hr.
At what time will they will be at distance of 52km apart from each other.
:
let t = time in hrs for the walker to be 52 km from the cyclist
and
(t-1) = time for the cyclist to be 52km from the walker
then
5t = the distance walked
and
16(t-1) = distance cycled
or we can write it as:
(16t-16)
:
Using pythag
%285t%29%5E2+%2B+%2816t-16%29%5E2+=+52%5E2
25t%5E2+%2B+256t%5E2+-+512t+%2B+256+=+2704
combine like terms to form a quadratic equation
281t%5E2+-+512t+%2B+256+-+2704+=+0
281t%5E2+-+512t+-+2448+=+0
Using the quadratic formula I got a positive solution of
t = 4 hrs, added to 7 Am, they were 52 km away at 11 AM
:
:
You can check this for yourself
Walker: 4*5 = 20 mi
Cyclist: 3*16 = 48 mi
dist = sqrt%2820%5E2%2B48%5E2%29