Liz rushes down onto a subway platform to find her train already departing. she stops and watches the cars go by. each car is 8.60 m long. the first moves past her in 1.80 s and the second in 1.66 s. find the constant acceleration of the train.

The average velocity can be calculated using the formula:

v = d / t

For the 1st car, the velocity is calculated as:

v1 = 8.60 m / 1.80 s = 4.78 m / s

While that of the 2nd car is:

v2 = 8.60 m / 1.66 s = 5.18 m / s

Now we can solve for the acceleration using the formula:

v2^2 = v1^2 + 2 a d

Rewriting in terms of a:

a = (v2^2 – v1^2) / 2 d

a = (5.18^2 – 4.78^2) / (2 * 8.6)

a = 0.23 m/s

Therefore the train has a constant acceleration of about 0.23 meters per second.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .