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
Yeah, your math is wrong. The right answer would be 13 seconds and 22 seconds.
Your math by definition is wrong, respond to me so I can show you the right answer.
To find the time(s) when your friend is 120 ft from you, you need to solve the equation d = 120. Let's substitute the given equation for d into the equation we need to solve:
|200 - 18t| = 120
To solve an absolute value equation, we need to consider both positive and negative solutions for the expression inside the absolute value.
First, let's consider the positive case:
200 - 18t = 120
Now, solve for t:
-18t = 120 - 200
-18t = -80
t = (-80)/(-18) = 40/9 ≈ 4.44 s
For the negative case, we set the expression inside the absolute value to its negative value:
200 - 18t = -120
Solve for t:
-18t = -120 - 200
-18t = -320
t = (-320)/(-18) = 320/18 ≈ 17.78 s
Therefore, your friend is approximately 120 ft away from you at t ≈ 4.44 s (positive case) and t ≈ 17.78 s (negative case).
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.