A car accelerates from rest at a constant rate of 2.2 m/s2 for 6.9 s. How far does the car travel in this time?

d = Vo*t + 0.5a*t^2,

d = 0 + 1.1*(6.9)^2 = 52.4m.

To find the distance traveled by the car, we can use the equation of motion:

distance = (initial velocity * time) + (0.5 * acceleration * time^2)

In this case, the car starts from rest, so the initial velocity is 0 m/s. The acceleration is given as 2.2 m/s^2, and the time is given as 6.9 s.

Plugging in these values into the equation, we get:

distance = (0 * 6.9) + (0.5 * 2.2 * 6.9^2)

First, let's calculate the term (0 * 6.9), which equals 0.

distance = 0 + (0.5 * 2.2 * 6.9^2)

Now, let's calculate the term 0.5 * 2.2 * 6.9^2:

distance = 0 + (0.5 * 2.2 * 47.61)

Multiplying 2.2 by 47.61, we get:

distance = 0 + (1.1 * 47.61)

Next, multiplying 1.1 by 47.61, we get:

distance = 0 + 52.371

Finally, adding 0 and 52.371, we get:

distance = 52.371

Therefore, the car travels a distance of 52.371 meters in 6.9 seconds when accelerating at a constant rate of 2.2 m/s^2.