A jet plane is flying at a constant altitude. At time t1=0t 1=0, it has components of velocity vx=90m/s, vy=110m/sv x = 90m/s, v y=110m/s. At time t2=30.0st 2=30.0s, the components are vx=−170m/s, vy=40m/sv x =−170m/s, v y=40m/s. (a) Sketch the velocity vectors at t1and t2.
How do these two vectors differ? For this time interval calculate
(b) the components of the average acceleration, and
(c) the magnitude and direction of the average acceleration.

The average acceleration over some time interval is equal to the ratio of the change in velocity over the duration of the interval , or

which can be split into the and components as

The magnitude of this average acceleration is

and its direction is such that

which corresponds to a direction of about 15.1º South of West.

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.

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 .