A thief was given a head-start of 15 hour. The velocity of the thief being 4 km/hr and the police chasing after him be 5 Km/hr. A dog is moving to and fro between the police and the thief, starting from the police at a velocity of 10 Km/hr. Every time the dog touches the thief it gives him a bite which reduces the speed of the thief by 10%.
- When will the police catch the thief?
- What would be the distance covered by the dog by that time?
- What would be the distance covered by the dog in the forward direction?
Answer Not given.
How can I get the answer without writing a program?
In a $(t,y)$ coordinate system, let $$(t_n,p_n)\qquad (n\geq0)$$ with $(t_0,p_0)=(0,0)$ be the "world points" where the dog meets the police, and let $$(h_n,s_n)\qquad(n\geq1)$$ be the "world points" where the dog hits the thief. Then $$p_n=5t_n\qquad (n\geq0)\ .\tag{1}$$ When the thief is hit for the first time we have $$60+4h_1=s_1=10 h_1\ ,$$ which leads to $(h_1,s_1)=(10,100)$. The sequel is governed by the following recursion formulas: $$\eqalign{s_n-p_n&=10(t_n-h_n)\cr s_{n+1}-p_n&=10(h_{n+1}-t_n)\cr s_{n+1}-s_n&=4\cdot 0.9^n(h_{n+1}-h_n)\ .\cr}\qquad(n\geq1)\tag{2}$$ Use $(1)$ in order to eliminate $p_n$ from $(2)$; then eliminate $t_n$. And on, and on$\ldots$