Rate/distance probs?

Question

I need help on how to solve these and show my work!

At Noon a horse and buggy headed north at 8 mph. two hours later the roadster headed south from the same location at 60 mph. At what time will the horse and buggy be 118 miles from the roadster?

Answer 1

start here -- After the first two hours, the horse and roadster are already 16 miles apart. The roadster did not travel any distance in this time, while the horse moved north at 8mph.

distance = rate x time = 8mph x 2 hrs = 16 miles

Next, set up your equation to take into account only the remaining distance that the two object need to move to achieve the desired 118 miles apart (this would be 118 miles - 16 miles = 102 miles)

(rate of the horse * t) + (rate of the roadster * t) = 102 miles

(8mph * t) + (60mph * t) = 102 miles

remove the mph and miles for clarity and get

8t + 60t = 102
68t = 102
t = 1.5 hours

Be careful....add this to the initial two hours that the buggy moved while the roadster did not and get

1.5 hours + 2.0 hours = 3.5 hours

Answer 2

This is very easy to do in your head! In two hours, when the roadster starts, the horse has already gone 16 miles. After 3 hours (3:00 pm) the horse has gone another 8 miles and the roadster has gone 60 miles, thus they are 16 + 8 + 60 = 84 miles apart. By 4:00 pm, the horse has gone another 8 miles for a total of 32 miles and the roadster has traveled a total of 120 miles, so they would be 152 miles apart. So back up to 3:30 pm or 3.5 elapsed hours. In 3.5 hrs the horse has gone 28 miles (8+8+8+4) and the roadster has gone 90 (60+30) miles so they would be 28 + 90 or 118 miles apart.

Answer 3

8mph
^
I
H R
I
V
60mph

after 2 hrs, horse 16 miles apart, separate (8+16)=24 mph
thus remaining 102 miles, 24mph
time is 102/24 = or .4.25hours
total time is 2hrs+4.25 hours = 6.25 hours
from noon, 6:25pm.
every hour after that