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 of stopping a car with given distances and initial speeds, we will break it down into steps.

### Given Data:
1. Distance covered when initial speed is 80.5 km/h: \( S_1 = 56.7 \, \text{m} \)
2. Distance covered when initial speed is 48.3 km/h: \( S_2 = 24.4 \, \text{m} \)
3. Initial speed 1: \( v_1 = 80.5 \, \text{km/h} \)
4. Initial speed 2: \( v_2 = 48.3 \, \text{km/h} \)

### Step 1: Convert speeds from km/h to m/s
To convert km/h to m/s, we use the conversion factor \( \frac{5}{18} \).

- For \( v_1 \):
\[ v_1 = 80.5 \times \frac{5}{18} = 22.36 \, \text{m/s} \]

- For \( v_2 \):
\[ v_2 = 48.3 \times \frac{5}{18} = 13.42 \, \text{m/s} \]

### Step 2: Set up equations for both scenarios
Let \( t_0 \) be the reaction time and \( a \) be the deceleration.

For the first case (initial speed \( v_1 \)):
\[ S_1 = v_1 t_0 + \frac{1}{2} a t^2 \]
Where \( t = \frac{v_1}{a} \) (time taken to stop after braking starts).

Substituting:
\[ 56.7 = 22.36 t_0 + \frac{1}{2} a \left(\frac{22.36}{a}\right)^2 \]
This simplifies to:
\[ 56.7 = 22.36 t_0 + \frac{22.36^2}{2a} \]

For the second case (initial speed \( v_2 \)):
\[ S_2 = v_2 t_0 + \frac{1}{2} a t^2 \]
Where \( t = \frac{v_2}{a} \).

Substituting:
\[ 24.4 = 13.42 t_0 + \frac{1}{2} a \left(\frac{13.42}{a}\right)^2 \]
This simplifies to:
\[ 24.4 = 13.42 t_0 + \frac{13.42^2}{2a} \]

### Step 3: Solve the equations
We now have two equations:
1. \( 56.7 = 22.36 t_0 + \frac{22.36^2}{2a} \) (Equation 1)
2. \( 24.4 = 13.42 t_0 + \frac{13.42^2}{2a} \) (Equation 2)

From Equation 1:
\[ 56.7 - 22.36 t_0 = \frac{22.36^2}{2a} \]

From Equation 2:
\[ 24.4 - 13.42 t_0 = \frac{13.42^2}{2a} \]

We can express both equations in terms of \( a \):
\[ a = \frac{22.36^2}{2(56.7 - 22.36 t_0)} \quad (1) \]
\[ a = \frac{13.42^2}{2(24.4 - 13.42 t_0)} \quad (2) \]

Setting these equal to each other:
\[ \frac{22.36^2}{2(56.7 - 22.36 t_0)} = \frac{13.42^2}{2(24.4 - 13.42 t_0)} \]

Cross-multiplying and simplifying will yield a linear equation in \( t_0 \). Solving this will give us the reaction time \( t_0 \).

### Step 4: Calculate deceleration \( a \)
Once we have \( t_0 \), we can substitute it back into either equation to find \( a \).

### Final Results:
(a) Reaction time \( t_0 \) is found to be approximately \( 0.74 \, \text{s} \).

(b) The magnitude of deceleration \( a \) is found to be approximately \( 6.2 \, \text{m/s}^2 \).

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