consider that the rider will be 60 feet away both approaching and receding. So, you want
|105-35t| = 60
for positive distance (still approaching),
105-35t = 60
t = 9/7 = 1.28
Going away, you have
105-35t = -60
t = 33/7 = 4.71
http://www.wolframalpha.com/input/?i=%7C105-35t%7C+%3D+60
Starting from 105 feet away, a person on a bicycle rides towards a checkpoint and then passes it. The rider is traveling at a constant rate of 35 feet per second. The distance between the bicycle and the checkpoint is given by the equation d = 105 - 35t. At what times is the bike 60 feet away from the checkpoint?
A: 1.3 sec and 2.6 sec
B: 1.3 sec and 4.7 sec*
C: 1.1 sec and 2.6 sec
D: 4.7 sec and 9.4 sec
I honestly took a wild guess since I have no idea to start. Any feedback would be insanely helpful and needed. Thank you!
2 answers
Anyone have all the answers
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