https://file.aiquickdraw.com/mathbot/file/1728764961807fruzvey2.jpg

Let's break down the problem from the image:

### Problem Recap
- A competitor moves in a straight line past three checkpoints \( P \), \( Q \), and \( R \). 
- **Distances**: \( PQ = 2.4 \, \text{km} \) and \( QR = 11.5 \, \text{km} \).  
- The competitor moves with **constant acceleration**.
- **Time**:
  - From \( P \) to \( Q \): \( 1 \, \text{hour} \).
  - From \( Q \) to \( R \): \( 1.5 \, \text{hours} \).

The question asks for the **acceleration of the competitor**.

---

### Assumptions and Equations

We assume the competitor’s motion follows **kinematic equations** since the acceleration is constant. Let’s denote:
- \( u \): Initial velocity at \( P \).
- \( a \): Constant acceleration (which we need to find).
- \( s_{PQ} = 2.4 \, \text{km} \) and \( s_{QR} = 11.5 \, \text{km} \) are the displacements.
- \( t_{PQ} = 1 \, \text{hr} \) and \( t_{QR} = 1.5 \, \text{hrs} \) are the times.

---

### Step-by-Step Solution

#### 1. Motion from \( P \) to \( Q \):
Using the kinematic equation for displacement:
\[
s_{PQ} = u \cdot t_{PQ} + \frac{1}{2} a \cdot t_{PQ}^2
\]
Substitute \( s_{PQ} = 2.4 \, \text{km} \) and \( t_{PQ} = 1 \, \text{hr} \):
\[
2.4 = u \cdot 1 + \frac{1}{2} a \cdot 1^2 \quad \Rightarrow \quad 2.4 = u + \frac{1}{2} a \tag{1}
\]

#### 2. Motion from \( Q \) to \( R \):
Using the same kinematic equation:
\[
s_{QR} = (u + a \cdot t_{PQ}) \cdot t_{QR} + \frac{1}{2} a \cdot t_{QR}^2
\]
The initial velocity for this segment is \( u + a \cdot t_{PQ} \) (since velocity increased by \( a \cdot t_{PQ} \) from \( P \) to \( Q \)).

Substitute \( s_{QR} = 11.5 \, \text{km} \), \( t_{QR} = 1.5 \, \text{hrs} \), and simplify:
\[
11.5 = (u + a \cdot 1) \cdot 1.5 + \frac{1}{2} a \cdot (1.5)^2
\]
\[
11.5 = 1.5 u + 1.5 a + \frac{1}{2} a \cdot 2.25
\]
\[
11.5 = 1.5 u + 1.5 a + 1.125 a \quad \Rightarrow \quad 11.5 = 1.5 u + 2.625 a \tag{2}
\]

---

#### 3. Solving the System of Equations

From Equation (1):
\[
u = 2.4 - \frac{1}{2} a \tag{3}
\]
Substitute Equation (3) into Equation (2):
\[
11.5 = 1.5 \left( 2.4 - \frac{1}{2} a \right) + 2.625 a
\]
Expand:
\[
11.5 = 3.6 - 0.75 a + 2.625 a
\]
\[
11.5 - 3.6 = 1.875 a
\]
\[
7.9 = 1.875 a
\]
\[
a = \frac{7.9}{1.875} = 4.21 \, \text{km/hr}^2
\]

---

### Final Answer

The acceleration of the competitor is:
\[
\boxed{4.21 \, \text{km/hr}^2}
\]

---

### Next Steps or Questions?
1. How would the solution change if the competitor had deceleration instead?
2. What is the total displacement from \( P \) to \( R \)?
3. Can you determine the competitor’s velocity when she reaches point \( R \)?
4. How would the answer differ if the time intervals were given in minutes instead of hours?
5. Could we solve this using energy principles (like kinetic energy)?  

**Tip:** Always ensure that your units are consistent (e.g., hours to seconds) when solving physics problems involving kinematics.

In an orienteering competition, a competitor moves in a straight line past three checkpoints, P, Q, and R, where PQ = 2.4 km and QR = 11.5 km. The competitor is modeled as a particle moving with constant acceleration. She takes 1 hour to travel from P to Q and 1.5 hours to travel from Q to R. Find: a) the acceleration of the competitor.

This problem involves calculating the acceleration of a competitor moving between checkpoints P, Q, and R in an orienteering competition. Given distances and time intervals, we use kinematic equations to determine the acceleration. Suitable for students in Grades 10-12 studying physics and motion.

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