SOLUTION: can anyone help me? A walker walks a course at a constant speed of 4 mph. 40 min after the walker begins, a cyclist starts on the same course at a constant speed of 16 mph. how

Algebra ->  Linear-equations -> SOLUTION: can anyone help me? A walker walks a course at a constant speed of 4 mph. 40 min after the walker begins, a cyclist starts on the same course at a constant speed of 16 mph. how       Log On


   

Question 9734: can anyone help me?
A walker walks a course at a constant speed of 4 mph. 40 min after the walker begins, a cyclist starts on the same course at a constant speed of 16 mph. how long after the cyclist begins does the cyclist overtake the walker? What distance have they traveled? Has teh walker completed the course before the cyclist catches up?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
For the walker:
d1+=+r1%2At1 r1 = 4 mph
d1+=+4%2At1 t1 = t2 + 2/3 hrs. (40 mins = 2/3 hrs)
For the cyclist:
d2+=+r2%2At2 r2 = 16 mph
d2+=+16%2At2
d1+=+4%28t2%2B2%2F3%29
When they meet, d1 = d2.
4%28t2%2B2%2F3%29+=+16%2At2 Simplify and solve for t2
4%2At2+%2B+8%2F3+=+16%2At2 Subtract 4*t2 from both sides.
8%2F3+=+12%2At2 Divide both sides by 12.
8%2F36+=+t2
t2 = 2/9 hours They will meet 2/9 hrs (13.3... mins) after the cyclist starts.
They will have traveled a distance of d1 = 16*t2 or 16 mph(2/9 hrs) = 3.56 miles.
Has the walker completed the course? Who knows! How long is the course?
Check:
d1+=+4+mph%28t1%29
d1+=+4+mph%28t2%2B2%2F3%29
d1+=+4%282%2F9+%2B+2%2F3%29
d1+=+4%282%2F9+%2B+6%2F9%29
d1+=+4%288%2F9%29
d1+=+32%2F9 miles.
d2+=+16+mph%28t2%29
d2+=+16+mph%282%2F9+hrs%29
d2+=+32%2F9 miles.