https://file.aiquickdraw.com/mathbot/file/172772516058373bzanem.jpg

Let's focus on solving part **(b)(iii)** of the problem from the image. Here is the breakdown of what needs to be done:

### Problem Outline for Part (b)(iii):
1. **Given Information:**
   - The car starts catching up with the bus when the bus begins to decelerate uniformly at 0.2 m/s².
   - The car takes **7 seconds** to catch up with the bus.
   - The distance between the car and the bus initially is **240 meters**.
   - The bus is traveling at **20 m/s** before deceleration.
   
2. **Goal:**
   - Show that the equation for the time \( T \) taken for the car to catch up is:
     \[
     T^2 - 50T - 2400 = 0
     \]
   - Find the speed of the bus when the car catches up with it.

---

### Step 1: Distance traveled by the car
We are given that the car catches up in 7 seconds. Let's denote \( T = 7 \) seconds.

- The car travels at a constant speed of 15 m/s, so the distance the car travels in this time is:
  \[
  d_{\text{car}} = 15T = 15 \times 7 = 105 \, \text{meters}.
  \]

### Step 2: Distance traveled by the bus
The bus decelerates from 20 m/s at a rate of 0.2 m/s². The equation of motion for the bus during deceleration is:
\[
v = u + at,
\]
where:
- \( u = 20 \, \text{m/s} \) (initial speed of the bus),
- \( a = -0.2 \, \text{m/s}^2 \) (negative because the bus is decelerating),
- \( t = T \, \text{seconds} \) (time of deceleration, which is 7 seconds).

So, the velocity of the bus after time \( T = 7 \) seconds is:
\[
v = 20 - 0.2T = 20 - 0.2 \times 7 = 20 - 1.4 = 18.6 \, \text{m/s}.
\]

### Step 3: Equation for distance difference
We are also given that the total distance between the car and the bus is 240 meters initially. For the car to catch up with the bus, the total distance traveled by the car plus this 240 meters should equal the total distance traveled by the bus during deceleration. 

So, the distance traveled by the bus while decelerating is given by the equation:
\[
d_{\text{bus}} = uT + \frac{1}{2} a T^2,
\]
where \( u = 20 \, \text{m/s} \), \( a = -0.2 \, \text{m/s}^2 \), and \( T = 7 \, \text{seconds} \).

Substituting into the equation for \( d_{\text{bus}} \):
\[
d_{\text{bus}} = 20T + \frac{1}{2}(-0.2)T^2 = 20T - 0.1T^2.
\]

Now, the total distance the car travels plus 240 meters should equal the distance traveled by the bus, so:
\[
240 + 15T = 20T - 0.1T^2.
\]

### Step 4: Rearranging the equation
Rearranging this equation to form a quadratic:
\[
240 + 15T = 20T - 0.1T^2
\]
\[
240 = 5T - 0.1T^2
\]
\[
0.1T^2 - 5T - 240 = 0.
\]
Multiplying through by 10 to eliminate decimals:
\[
T^2 - 50T - 2400 = 0,
\]
which is the required equation.

---

### Step 5: Solving for the bus's speed when the car catches up
Now, we want to find the speed of the bus when the car catches up. This happens at \( T = 7 \) seconds.

The speed of the bus is given by:
\[
v_{\text{bus}} = u + aT = 20 + (-0.2)(7) = 20 - 1.4 = 18.6 \, \text{m/s}.
\]

Thus, the speed of the bus when the car catches up is **18.6 m/s**.

---

### Final Answer:
- The quadratic equation is \( T^2 - 50T - 2400 = 0 \).
- The speed of the bus when the car catches up is **18.6 m/s**.

Would you like more details on any part of this solution?

---

Here are five related questions to think about:
1. How would the time change if the car's speed was slightly faster?
2. How would you solve the quadratic equation \( T^2 - 50T - 2400 = 0 \) using the quadratic formula?
3. What happens if the bus decelerates at a different rate? How would that affect the outcome?
4. How would you calculate the total distance traveled by both vehicles by the time they meet?
5. What if the bus did not decelerate—how would the problem change?

**Tip:** Always verify your units when solving physics problems to ensure consistency across equations.

Solve a kinematics-based quadratic equation involving a car catching up to a bus. The problem involves uniform acceleration and deceleration, applying algebraic and motion formulas to derive the equation T^2 - 50T - 2400 = 0. This problem is suited for students in Grades 11-12.

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