A car traveling at a constant speed of 33 m/s passes a trooper hidden behind a billboard. One second later the trooper starts the car with a constant acceleration of 2.62 m/s^2.

How long after the trooper starts the chase does he overtake the speeding car? (answer in units of s)

d1 = Vo*(t+1) + 0.5at^2,

d1 = 33(t+1) + 0.5*0*(t+1)^2,
d1 = 33(t+1) + 0 = 33t + 33.

d2 = 0 + 0.5*2.62*t^2,
d2 = 1.31t^2.

d1 = d2,
33t + 33 = 1.31t^2,
1.31t^2 - 33t - 33 = 0,
Solve using Quadratic Formula and get:
t = 26.15s, and -0.96s.
Use the positive valuie:
t = 26.15s.

To solve this problem, we need to find the time it takes for the trooper's car to catch up with the speeding car. Let's break down the steps to solve this problem:

Step 1: Calculate the distance covered by the speeding car during the one-second delay.
To do this, we need to find the distance traveled by the speeding car in that one second.
Distance = Speed × Time
Distance = 33 m/s × 1 s = 33 m

Step 2: Use kinematic equations to solve for the time it takes for the trooper's car to overtake the speeding car.
We can use the equation: Distance = Initial velocity × Time + (1/2) × Acceleration × Time^2
In this case, the initial velocity of the trooper's car is 0 m/s (since it starts from rest) and the acceleration is 2.62 m/s^2. The distance covered by the trooper's car is the same as the distance covered by the speeding car, which is 33 m.
Thus, the equation becomes: 33 m = 0 m/s × t + (1/2) × 2.62 m/s^2 × t^2
Simplifying the equation: 33 m = (1.31) × t^2
Dividing both sides of the equation by 1.31: t^2 = 25.19
Taking the square root of both sides: t ≈ 5.019 s

Step 3: Round the answer to the appropriate significant figures.
Since the initial speeds and distances are given with two significant figures, we round the final answer to the same number of significant figures.
Therefore, the trooper overtakes the speeding car approximately 5.02 seconds after starting the chase.