Question
Starting from 200 ft away, your friend rides his bike toward you and then passes by you at a speed of 18 ft/s. His distance d (in feet) from you t seconds after he started riding his bike is given by d=|200-18t |. At what time(s) is he 120 ft from you?
I got t=4 4/9s but not sure how to get second time?
Thank you
I got t=4 4/9s but not sure how to get second time?
Thank you
Answers
you want
|200-18t| = 120
For 200-18t >= 0, you have
200-18t = 120
18t = 80
t = 4 4/9
For t>11.11, 200-18t < 0, so
-(200-18t) = 120
-200+18t = 120
18t = 320
t = 17 7/9
So, 4 4/9 seconds after starting, he is 120 ft away. Then he passes you at 11 1/9 seconds, and at 17 7/9 seconds he is 120 ft past you.
Note that 11 1/9 - 4 4/9 = 17 7/9 - 11 1/9
That's because it takes 120/18 = 6 6/9 seconds to go 120 ft.
|200-18t| = 120
For 200-18t >= 0, you have
200-18t = 120
18t = 80
t = 4 4/9
For t>11.11, 200-18t < 0, so
-(200-18t) = 120
-200+18t = 120
18t = 320
t = 17 7/9
So, 4 4/9 seconds after starting, he is 120 ft away. Then he passes you at 11 1/9 seconds, and at 17 7/9 seconds he is 120 ft past you.
Note that 11 1/9 - 4 4/9 = 17 7/9 - 11 1/9
That's because it takes 120/18 = 6 6/9 seconds to go 120 ft.
Your math by definition is wrong, respond to me so I can show you the right answer.
Yeah, your math is wrong. The right answer would be 13 seconds and 22 seconds.