To stop a car, first you require a certain reaction time to begin braking, then the car slows under the constant braking deceleration. Suppose that the total distance moved by your car during these two phases is 56.7 m when its initial Speed is 80.5 km//h and 24.4 m when its initial speed in 48.3 km//h. What are
(a) your reaction time and
(b) the magnitude or the deceleration?

Explanation

To solve the problem, we need to find the reaction time (t₀) and the magnitude of deceleration (a) for the car. We will use the equations of motion and the information provided in the question.

### Step-by-Step Solution:

1. **Convert Speeds from km/h to m/s:**
- The initial speed of the car (V₁) is 80.5 km/h.
- To convert km/h to m/s, use the conversion factor: \( \text{m/s} = \text{km/h} \times \frac{5}{18} \).
- Therefore,
\[ V₁ = 80.5 \times \frac{5}{18} = 22.4 \, \text{m/s} \]
- The initial speed of the car (V₂) is 48.3 km/h.
- Similarly,
\[ V₂ = 48.3 \times \frac{5}{18} = 13.42 \, \text{m/s} \]

2. **Set Up the Equations of Motion:**
- For the first scenario (initial speed V₁ and distance 56.7 m):
\[ S₁ = V₁ t₀ + \frac{1}{2} a t₀^2 \]
\[ 56.7 = 22.4 t₀ + \frac{1}{2} a t₀^2 \quad \text{(Equation 1)} \]
- For the second scenario (initial speed V₂ and distance 24.4 m):
\[ S₂ = V₂ t₀ + \frac{1}{2} a t₀^2 \]
\[ 24.4 = 13.42 t₀ + \frac{1}{2} a t₀^2 \quad \text{(Equation 2)} \]

3. **Subtract Equation 2 from Equation 1:**
- Subtracting the second equation from the first gives:
\[ 56.7 - 24.4 = (22.4 - 13.42) t₀ + \left(\frac{1}{2} a t₀^2 - \frac{1}{2} a t₀^2\right) \]
\[ 32.3 = 9.0 t₀ \]
- Solving for \( t₀ \):
\[ t₀ = \frac{32.3}{9.0} \approx 3.58 \, \text{seconds} \]

4. **Substitute t₀ back to find a:**
- Substitute \( t₀ \) back into either equation to find \( a \). Using Equation 1:
\[ 56.7 = 22.4 \times 3.58 + \frac{1}{2} a (3.58)^2 \]
- Calculate \( 22.4 \times 3.58 \):
\[ 22.4 \times 3.58 \approx 80.032 \]
- Therefore,
\[ 56.7 = 80.032 + \frac{1}{2} a (12.8164) \]
- Rearranging gives:
\[ 56.7 - 80.032 = \frac{1}{2} a (12.8164) \]
\[ -23.362 = \frac{1}{2} a (12.8164) \]
- Solving for \( a \):
\[ a = \frac{-23.362 \times 2}{12.8164} \approx -3.64 \, \text{m/s}^2 \]

### Final Answers:
(a) The reaction time \( t₀ \) is approximately **3.58 seconds**.
(b) The magnitude of the deceleration \( a \) is approximately **3.64 m/s²**.