When a deer was 48 m from a leopard, the leopard starts chasing the deer and the deer immediately st

Q: Why do we check the separation exactly at $t=6$ seconds?

The separation between them decreases as long as the leopard is faster than the deer. It starts increasing once the leopard becomes slower. If we assume $v_D$ is around 17 m/s, the leopard's speed drops from 20 m/s to 15 m/s at $t=6$ s. This is the exact moment the leopard goes from being faster to slower than the deer, making it the point of minimum separation.

Q: What if the deer gets caught before 6 seconds?

We can verify this! At $t=4$ s, the leopard has covered 110 m. The deer, running at 17 m/s, is at $48 + 17(4) = 116$ m. The deer is still 6 meters ahead. It only gets caught (separation becomes zero) exactly at $t=6$ s.

Q: Why can't we just equate their average speeds?

Average speed doesn't account for the instantaneous positions. The leopard could catch the deer early on while its speed is very high, even if its overall average speed later drops. We must ensure the deer is ahead at *every* instant.

Q: Is the leopard's acceleration constant?

No, the problem states the speed reduces in 'equal steps of 5 m/s after every 2 s interval'. This means the speed is constant for 2 seconds, then abruptly drops. It's a step function, not uniform deceleration.

Q: What if the deer's speed was 16 m/s?

If $v_D = 16$ m/s, at $t=4$ s, the deer is at $48 + 64 = 112$ m, and the leopard is at 110 m. In the next interval, the leopard runs at 20 m/s. The relative speed is $20 - 16 = 4$ m/s. The 2-meter gap will be closed in just 0.5 seconds. The deer gets caught at $t=4.5$ s!

Question

When a deer was 48 m from a leopard, the leopard starts chasing the deer and the deer immediately starts running away from the leopard with constant velocity. A leopard cannot run at high speeds for a long time and has to slow down due to fatigue. If we assume that the leopard starts with an initial speed of 30 m/s and reduces its speed in equal steps of 5 m/s after every 2 s interval, at what minimum speed must the deer run to escape from the leopard?

Sigma Percentile · Accepted Answer

### The Anatomy of the Chase

Imagine a high-stakes chase on the savanna. A deer spots a leopard and immediately bolts. The deer has a $48 	ext{ m}$ head start and runs at a constant velocity, $v_D$. The leopard, relying on explosive power, bursts into a sprint at $30 	ext{ m/s}$. However, it cannot maintain this blistering pace. Fatigue sets in, and its speed drops by $5 	ext{ m/s}$ every $2 	ext{ s}$.

This means the leopard's speed is a step function: $30 	ext{ m/s}$ for the first two seconds, $25 	ext{ m/s}$ for the next two, $20 	ext{ m/s}$ for the next, and so on. We need to find the absolute minimum speed the deer must maintain to ensure it never gets caught.

### The Escape Condition

For the deer to successfully escape, its position must always be strictly ahead of the leopard's position. Mathematically, the distance covered by the deer, plus its initial $48 	ext{ m}$ advantage, must be greater than or equal to the distance covered by the leopard at any given time $t$:

$$S_D(t) + 48 \ge S_L(t)$$

Since the deer runs at a constant speed, $S_D(t) = v_D t$. The inequality becomes:

$$v_D t + 48 \ge S_L(t)$$

### Tracking the Predator

Let's calculate the exact distance the leopard covers at the end of each $2$-second interval.

At $t = 2 	ext{ s}$, running at $30 	ext{ m/s}$, it covers:
$$S_L(2) = 30 	imes 2 = 60 	ext{ m}$$

At $t = 4 	ext{ s}$, its speed was $25 	ext{ m/s}$ for the last two seconds:
$$S_L(4) = 60 + (25 	imes 2) = 110 	ext{ m}$$

At $t = 6 	ext{ s}$, its speed was $20 	ext{ m/s}$:
$$S_L(6) = 110 + (20 	imes 2) = 150 	ext{ m}$$

At $t = 8 	ext{ s}$, its speed was $15 	ext{ m/s}$:
$$S_L(8) = 150 + (15 	imes 2) = 180 	ext{ m}$$

### The Critical Moment of Maximum Danger

When is the deer in the most danger? The gap between the two animals shrinks as long as the leopard is running faster than the deer ($v_L > v_D$). The moment the leopard's speed drops below the deer's speed, the gap will start widening again. Therefore, the minimum separation occurs exactly when the relative velocity crosses zero.

Let's assume this critical moment happens at $t = 6 	ext{ s}$. Right before $t = 6 	ext{ s}$, the leopard is running at $20 	ext{ m/s}$. Right after, it drops to $15 	ext{ m/s}$. If the deer's speed is somewhere between $15 	ext{ m/s}$ and $20 	ext{ m/s}$, then $t = 6 	ext{ s}$ is indeed the point of minimum separation.

### The Final Calculation

Let's enforce our escape condition at the critical time $t = 6 	ext{ s}$. The deer's position must be at least equal to the leopard's position:

$$v_D(6) + 48 \ge S_L(6)$$

Substituting the leopard's distance at $6$ seconds:

$$6 v_D + 48 \ge 150$$

Subtracting $48$ from both sides:

$$6 v_D \ge 102$$

Dividing by $6$:

$$v_D \ge 17 	ext{ m/s}$$

### Verification

Let's verify our assumption. If the deer runs at exactly $17 	ext{ m/s}$, does it ever get caught before $t = 6 	ext{ s}$?

Let's check $t = 4 	ext{ s}$. The leopard is at $110 	ext{ m}$. The deer is at $48 + 17(4) = 48 + 68 = 116 	ext{ m}$. The deer is still $6 	ext{ m}$ ahead!

Between $t = 4 	ext{ s}$ and $t = 6 	ext{ s}$, the leopard runs at $20 	ext{ m/s}$, which is $3 	ext{ m/s}$ faster than the deer. Over these $2$ seconds, the leopard closes the $6 	ext{ m}$ gap exactly, bringing the separation to zero at $t = 6 	ext{ s}$. After $t = 6 	ext{ s}$, the leopard slows to $15 	ext{ m/s}$, and the deer (at $17 	ext{ m/s}$) begins to pull away.

Our logic holds perfectly. The minimum speed required for the deer to escape is **$17 	ext{ m/s}$**.

Enter Numerical Value:

Visualized Solution

Initial Setup

Condition for Escape

Leopard's Distance } S_L(t)

Finding the Critical Time

Calculating Minimum } v_D

Verification

The Sigma Insight: Relative Velocity

The Anatomy of the Chase

The Escape Condition

Tracking the Predator

The Critical Moment of Maximum Danger

The Final Calculation

Verification

Similar Questions

Comprehension Passage

When and for how long can the particle C fulfil the given condition?

What is speed of the particle C at the instant $t = 5 s$ ?

What is modulus of acceleration of the particle C at the instant $t = 5 s$ ?

At the instant, when the line joining locations of A and B is perpendicular to the line joining locations of B and C, what are the magnitudes of velocities of C relative to A and B respectively?

A boat which has a speed of $5 km/h$ in still water crosses a river of width $1 km$ along the shortest possible path in $15 min$ . The velocity of the river water in km/h is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Escape .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Leopard's Distance } S_L(t) .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Critical Time .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Minimum } v_D .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Verification .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

The Sigma Insight: Relative Velocity

The Anatomy of the Chase

The Escape Condition

Tracking the Predator

The Critical Moment of Maximum Danger

The Final Calculation

Verification

Similar Questions

Comprehension Passage

When and for how long can the particle C fulfil the given condition?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What is speed of the particle C at the instant t=5 s?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What is modulus of acceleration of the particle C at the instant t=5 s?

At the instant, when the line joining locations of A and B is perpendicular to the line joining locations of B and C, what are the magnitudes of velocities of C relative to A and B respectively?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest possible path in 15 min. The velocity of the river water in km/h is

Initial Setup

Condition for Escape

Leopard's Distance } S_L(t)

Finding the Critical Time

Calculating Minimum } v_D

Verification

What is speed of the particle C at the instant $t = 5 s$ ?

What is modulus of acceleration of the particle C at the instant $t = 5 s$ ?

A boat which has a speed of $5 km/h$ in still water crosses a river of width $1 km$ along the shortest possible path in $15 min$ . The velocity of the river water in km/h is