kichu and sachu were on a morning walk on teaching a junction kichu moved exactly east and sachu moved exactly north wards sachu was at a  speed of 30metres per minute than kichu.after 10 minutes the shortest distance between them was 11.5kilometer.draw a rough figure showing the position of kichu and sachu and the junction.calculate the distance Covered by each.find respective speed

Let's break down the problem step by step:

### Problem Breakdown:

1. **Direction of Movement:**
   - Kichu moved exactly east.
   - Sachu moved exactly north.

2. **Speed Information:**
   - Sachu’s speed is 30 meters per minute faster than Kichu’s speed.

3. **Time:**
   - Both moved for 10 minutes.

4. **Distance Information:**
   - The shortest distance between them after 10 minutes is 11.5 kilometers (11,500 meters).

### Step 1: Let’s Define Variables

- Let \( v_k \) be Kichu’s speed in meters per minute.
- Then, Sachu’s speed is \( v_s = v_k + 30 \) meters per minute.

### Step 2: Calculate Distances

- Kichu’s distance covered in 10 minutes: \( d_k = 10 \times v_k \) meters.
- Sachu’s distance covered in 10 minutes: \( d_s = 10 \times (v_k + 30) \) meters.

### Step 3: Apply Pythagoras Theorem

Since Kichu and Sachu moved in perpendicular directions, the distance between them forms the hypotenuse of a right triangle:

\[
\text{Distance} = \sqrt{d_k^2 + d_s^2} = 11,500 \text{ meters}
\]

Substituting \( d_k \) and \( d_s \) into the equation:

\[
\sqrt{(10v_k)^2 + \left(10(v_k + 30)\right)^2} = 11,500
\]

### Step 4: Simplify and Solve the Equation

Square both sides:

\[
(10v_k)^2 + \left(10(v_k + 30)\right)^2 = 11,500^2
\]

Simplify:

\[
100v_k^2 + 100(v_k + 30)^2 = 132,250,000
\]

Divide through by 100:

\[
v_k^2 + (v_k + 30)^2 = 1,322,500
\]

Expand and simplify:

\[
v_k^2 + v_k^2 + 60v_k + 900 = 1,322,500
\]

\[
2v_k^2 + 60v_k + 900 = 1,322,500
\]

Subtract 900 from both sides:

\[
2v_k^2 + 60v_k = 1,321,600
\]

Divide through by 2:

\[
v_k^2 + 30v_k = 660,800
\]

This is a quadratic equation. To solve it, we can use the quadratic formula:

\[
v_k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \( a = 1 \), \( b = 30 \), and \( c = -660,800 \).

Let me calculate the values now.The possible solutions for Kichu's speed (\(v_k\)) are:

1. \( v_k \approx 798.03 \) meters per minute (positive and valid)
2. \( v_k \approx -828.03 \) meters per minute (negative and invalid)

So, Kichu's speed is approximately \( 798.03 \) meters per minute.

### Step 5: Calculate Sachu's Speed and Distances Covered

- Sachu's speed: \( v_s = v_k + 30 \approx 798.03 + 30 = 828.03 \) meters per minute.
- Distance covered by Kichu: \( d_k = 10 \times 798.03 = 7980.34 \) meters.
- Distance covered by Sachu: \( d_s = 10 \times 828.03 = 8280.34 \) meters.

### Final Answer:
- **Kichu’s Speed:** 798.03 meters per minute.
- **Sachu’s Speed:** 828.03 meters per minute.
- **Distance Covered by Kichu:** 7980.34 meters.
- **Distance Covered by Sachu:** 8280.34 meters.

### Rough Figure Description:
- Imagine a right-angled triangle where the horizontal side represents the distance covered by Kichu (7980.34 meters) to the east, and the vertical side represents the distance covered by Sachu (8280.34 meters) to the north.
- The hypotenuse of this triangle is the shortest distance between them, which is 11,500 meters.

Would you like any further details or have any questions?

Here are 5 related questions:
1. How does the speed difference affect the shortest distance between Kichu and Sachu?
2. What if Sachu's speed was 20 meters per minute faster than Kichu's? How would that change the result?
3. How would the distance change if they moved for 15 minutes instead of 10 minutes?
4. Can you calculate the time it would take for the shortest distance to reach 15 kilometers?
5. How would the calculation change if Kichu moved west instead of east?

**Tip:** When solving quadratic equations, always check for both the positive and negative roots, but remember that in physical problems, only positive values typically make sense.

Explore how to calculate the shortest distance between Kichu and Sachu using Pythagorean Theorem and solving quadratic equations. Understand the concept of relative speed and distance covered by each in a specific time frame. Suitable for high school students.

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