A car traveling at a constant speed of 33.4 m/s passes a trooper hidden behind a billboard. One second later the trooper starts the car with a constant acceleration of 3.28 m/s2 How long after the trooper starts the chase does he overtake the speeding car

To find out how long after the trooper starts the chase does he overtake the speeding car, we can follow these steps:

1. Find the initial velocity of the trooper's car:
- It is given that the trooper's car starts with a constant acceleration of 3.28 m/s^2, but we need to know the initial velocity.
- Since the car starts from rest, the initial velocity is 0 m/s.

2. Determine the position equation for both cars at any given time:
- For the trooper's car:
- The position of the trooper's car (s1) can be given by the equation s1 = 0.5 * a * t^2 + v0*t, where a is the acceleration (3.28 m/s^2), t is the time, and v0 is the initial velocity (0 m/s).
- For the speeding car:
- The position of the speeding car (s2) can be given by the equation s2 = v1 * t, where v1 is the constant speed of the car (33.4 m/s).

3. Set up an equation to determine the time when the two cars are at the same position:
- At the moment of overtaking, the positions of both cars will be equal. So, we can set s1 = s2 and solve for t.

0.5 * 3.28 * t^2 + 0*t = 33.4 * t

4. Solve the equation for t:
- Simplify the equation:
1.64 * t^2 = 33.4 * t
1.64 * t^2 - 33.4 * t = 0
- Divide both sides of the equation by t:
1.64 * t - 33.4 = 0
- Solve for t using the quadratic formula or factoring:
t ≈ (33.4 ± √(33.4^2 - 4*1.64*0)) / (2*1.64)
t ≈ (33.4 ± √(1115.56 - 0)) / 3.28
t ≈ (33.4 ± √1115.56) / 3.28

We only consider the positive solution, as the negative one doesn't make sense in this context:
t ≈ (33.4 + √1115.56) / 3.28
t ≈ (33.4 + 33.42) / 3.28
t ≈ 66.82 / 3.28
t ≈ 20.39 seconds

Therefore, the trooper will overtake the speeding car approximately 20.39 seconds after starting the chase.