As soon as the stop light changes from red to green, a car at rest at the intersection accelerates with a constant acceleration of 2.2 m/s/s. At the same moment, a truck passes by, moving at a constant speed of 9.5 m/s. How far beyond the intersection will the car catch the truck? How fast will the car be going at this point?

Dc = Dt

0.5a*t^2 = 9.5*t
1.1t^2 = 9.5t
1.1t^2 - 9.5t = 0
t = 8.64 s. (Use Quad. formula).
D = 9.5 * 8.64 = 82.1 m.

V = Vo + a*t = 0 + 2.2*8.64 = 19.0 m/s.

To find the distance the car will catch the truck beyond the intersection and the speed of the car at that point, we can use the concepts of kinematics.

First, let's define some variables:
- Let t be the time it takes for the car to catch the truck beyond the intersection.
- Let d be the distance the car travels beyond the intersection.
- Let v_car be the final velocity of the car when it catches the truck.
- Let v_truck be the constant speed of the truck.
- Let a be the constant acceleration of the car.

We need to find the values of d and v_car.

We know that the car's acceleration is 2.2 m/s^2, and the truck's speed is 9.5 m/s. Since they started moving at the same time, the car will need to catch up to the truck.

The relationship between distance, velocity, time, and acceleration is given by the equation:
d = v_0t + (1/2)at^2

Since the car starts from rest, its initial velocity (v_0) is 0. Plugging in these values into the equation, we get:
d = (1/2)at^2

We also know that the car's final velocity, v_car, is equal to the truck's speed, v_truck, when the car catches the truck:
v_car = v_truck

Now, let's find the time it takes for the car to catch up to the truck. We can use the kinematic equation for final velocity:
v_car = v_0 + a*t

Since the car starts from rest (v_0 = 0), we can simplify the equation to:
v_car = a*t

Now, equating v_car to v_truck, we get:
v_truck = a*t

From this equation, we can solve for t:
t = v_truck / a

Substituting the values for v_truck (9.5 m/s) and a (2.2 m/s^2), we find:
t = 9.5 m/s / 2.2 m/s^2
t ≈ 4.32 s

Now that we have found the time it takes for the car to catch the truck (t ≈ 4.32 s), we can find the distance the car travels beyond the intersection using the equation:
d = (1/2)at^2

Substituting the values for a (2.2 m/s^2) and t (4.32 s), we have:
d = (1/2)(2.2 m/s^2)(4.32 s)^2
d ≈ 40.1 m

So, the car will catch the truck approximately 40.1 meters beyond the intersection.

To find the final velocity of the car, we can use the equation:
v_car = a*t

Substituting the values for a (2.2 m/s^2) and t (4.32 s), we have:
v_car = (2.2 m/s^2)(4.32 s)
v_car ≈ 9.5 m/s

Therefore, the car will be going approximately 9.5 m/s at the point it catches up to the truck.