A car moving at a constant velocity of 46 m/s passes a traffic cop who is readily sitting on his motorcycle. After a reaction time of 1.0s, the cop begins to chase the spoeding car with a constant acceleration of 4.0 `m//s^(2)`. How much time does the cop then need to overtake the speeding car ?

Answer

To solve the problem step by step, we will analyze the motion of both the car and the cop.

### Step 1: Calculate the distance traveled by the car during the cop's reaction time.
The car is moving at a constant velocity of 46 m/s. The cop has a reaction time of 1.0 seconds before he starts chasing the car.

**Distance traveled by the car during the reaction time:**
\[ \text{Distance} = \text{Velocity} \times \text{Time} = 46 \, \text{m/s} \times 1.0 \, \text{s} = 46 \, \text{m} \]

### Step 2: Set up the equations of motion.
After the reaction time, the cop starts chasing the car with a constant acceleration of 4.0 m/s².

- **Distance traveled by the car after the reaction time (for time \( t \)):**
\[ d_{\text{car}} = 46 \, \text{m} + 46 \, \text{m/s} \times t \]
This accounts for the initial distance of 46 m and the distance traveled at 46 m/s for time \( t \).

- **Distance traveled by the cop (starting from rest with acceleration):**
\[ d_{\text{cop}} = \frac{1}{2} a t^2 = \frac{1}{2} \times 4 \, \text{m/s}^2 \times t^2 = 2t^2 \]

### Step 3: Set the distances equal to find the time when the cop overtakes the car.
To find the time \( t \) when the cop overtakes the car, we set the distances equal:
\[ 46 + 46t = 2t^2 \]

### Step 4: Rearrange the equation.
Rearranging gives:
\[ 2t^2 - 46t - 46 = 0 \]

### Step 5: Solve the quadratic equation.
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2 \), \( b = -46 \), and \( c = -46 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-46)^2 - 4 \times 2 \times (-46) = 2116 + 368 = 2484 \]

Now, applying the quadratic formula:
\[ t = \frac{46 \pm \sqrt{2484}}{4} \]

Calculating \( \sqrt{2484} \):
\[ \sqrt{2484} \approx 49.84 \]

Now substituting back:
\[ t = \frac{46 \pm 49.84}{4} \]

Calculating the two possible values for \( t \):
1. \( t = \frac{46 + 49.84}{4} \approx \frac{95.84}{4} \approx 23.96 \, \text{s} \)
2. \( t = \frac{46 - 49.84}{4} \) gives a negative value, which we discard since time cannot be negative.

### Final Answer:
The time needed for the cop to overtake the speeding car is approximately **24 seconds**.

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