SOLUTION: A boy at 8:00 A.M. started to walk at the rate of 4 km/hr for 2 hours and 45 minutes, after which a man follows to overtake him with a rate of 4.5 km/hr at the 1st hour, 4.75 km/h

Question 1172107: A boy at 8:00 A.M. started to walk at the rate of 4 km/hr for 2 hours and 45 minutes, after which
a man follows to overtake him with a rate of 4.5 km/hr at the 1st hour, 4.75 km/hr for the 2nd
hour, and so on, increasing his rate by a quarter of a km each hour. Find the time when the man
overtook the boy.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
oy at 8:00 A.M. started to walk at the rate of 4 km/hr for 2 hours and 45 minutes, after which a man follows to overtake him with a rate of 4.5 km/hr at the 1st hour, 4.75 km/hr for the 2nd hour, and so on, increasing his rate by a quarter of a km each hour.
Find the time when the man overtook the boy.
:
Assume this means that the man started after the boy had walked for 2 hr 45 min or at 10:45'
;
let t = no. of hrs the man walked when overtaken by the boy
then
(t+2.75) = no. hrs the boy walked
:
We have to get some idea of how long it will take in order to know the speeds that the man will walk.
At an average rate of 5 km/h; He will travel 40 km in 8 hrs
The boy has already walked 11 km when the man starts (2.75*4)
In 8 hrs he will have traveled 32 km plus the 11 km or 43 km
let's assume t will about 8 hrs
The man will walk 8 different speeds, each for 1 hr.
:
Write a distance equation, both walked the same dist. Dist = speed * time
4.5 + 4.75 + 5 + 5.25 + 5.5 + 5.75 + 6 + 6.25(t-7) = 4t + 11
37.25 + 6.25t - 43.75 = 4t + 11
6.25t - 6.5 = 4t + 11
6.25t - 4t + 11 + 6.5
2.25t = 17.5
t = 17.5/2.25
t = 7.78 hrs for man to catch the boy About: 7 + .78(60) = 7 hrs, 47
What time was it?
Starting at 10:45 add 7 hrs 47 min
10:45
7:47
-----
17 hrs 92 min or 18:32 or 6:32 PM The man overtook the boy
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