Physics Solution: Pursuit with Step Deceleration
Problem Analysis
A leopard chases a deer initially 48m away. The leopard’s speed decreases in steps of 5 m/s every 2 seconds, starting from 30 m/s. The deer moves at a constant velocity $v_D$. We need the minimum $v_D$ to ensure escape.
1. Leopard’s Displacement Profile
The leopard’s speed $v_L(t)$ is a step function.
- $0 \le t < 2$: $v = 30$. Dist $S_L(2) = 60$.
- $2 \le t < 4$: $v = 25$. Dist $S_L(4) = 60 + 50 = 110$.
- $4 \le t < 6$: $v = 20$. Dist $S_L(6) = 110 + 40 = 150$.
- $6 \le t < 8$: $v = 15$. Dist $S_L(8) = 150 + 30 = 180$.
2. Escape Condition
Problem Analysis
A leopard chases a deer initially 48m away. The leopard’s speed decreases in steps of 5 m/s every 2 seconds, starting from 30 m/s. The deer moves at a constant velocity $v_D$. We need the minimum $v_D$ to ensure escape.
1. Leopard’s Displacement Profile
The leopard’s speed $v_L(t)$ is a step function.
- $0 \le t < 2$: $v = 30$. Dist $S_L(2) = 60$.
- $2 \le t < 4$: $v = 25$. Dist $S_L(4) = 60 + 50 = 110$.
- $4 \le t < 6$: $v = 20$. Dist $S_L(6) = 110 + 40 = 150$.
- $6 \le t < 8$: $v = 15$. Dist $S_L(8) = 150 + 30 = 180$.
Let the deer’s position be $x_D(t) = 48 + v_D t$. Let the leopard’s position be $x_L(t) = S_L(t)$. For escape, we require $x_D(t) \ge x_L(t)$ for all $t$. $$ v_D \ge \frac{S_L(t) – 48}{t} $$ We must find the maximum value of the RHS over time $t$.
3. Checking Critical Points
We check the end of each interval ($t=2, 4, 6…$):
At t=2:
$$ v_{req} = \frac{60 – 48}{2} = \frac{12}{2} = 6 \text{ m/s} $$
At t=4:
$$ v_{req} = \frac{110 – 48}{4} = \frac{62}{4} = 15.5 \text{ m/s} $$
At t=6:
$$ v_{req} = \frac{150 – 48}{6} = \frac{102}{6} = 17 \text{ m/s} $$
At t=8:
$$ v_{req} = \frac{180 – 48}{8} = \frac{132}{8} = 16.5 \text{ m/s} $$
The requirement peaks at $t=6$ seconds with $v_D = 17$ m/s. If $v_D = 17$, the deer is exactly caught (gap closes to 0) at $t=6$ but then pulls away because at $t=6$, leopard speed drops to 15 m/s, which is slower than 17 m/s.
Answer: 17 m/s