SOLUTION: Cliff left point A at 8:00 A.M walking east at 3kph. Renz left point A at 9:00 A.M walking north at 4 kph. The distance between the two boys since 9:00 A.M may be expressed as d(t)

Question 1199939: Cliff left point A at 8:00 A.M walking east at 3kph. Renz left point A at 9:00 A.M walking north at 4 kph. The distance between the two boys since 9:00 A.M may be expressed as d(t)=√(〖at〗^2+2βt+β) where d and t denote the respective distance and time. Find a-β.
Answer by textot(100) (Show Source): You can put this solution on YOUR website!
**1. Determine the Distances Traveled**
* **Cliff:** Since Cliff started 1 hour earlier and walks at 3 kph, he has traveled 3 km east by 9:00 AM.
* **Renz:** Let 't' be the time in hours since 9:00 AM. Renz travels 4t kilometers north.
**2. Use Pythagorean Theorem**
* The distance between Cliff and Renz at time 't' can be found using the Pythagorean theorem:
d(t) = √[(Cliff's distance)² + (Renz's distance)²]
d(t) = √[(3 + 3t)² + (4t)²]
d(t) = √(9 + 18t + 9t² + 16t²)
d(t) = √(25t² + 18t + 9)
**3. Compare with the Given Expression**
* Compare the derived expression with the given expression:
* d(t) = √(at² + 2βt + β)
* d(t) = √(25t² + 18t + 9)
* We can see that:
* a = 25
* 2β = 18
* β = 9
**4. Calculate a - β**
* a - β = 25 - 9 = 16
**Therefore, a - β = 16.**

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