A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t) . At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.80 m/s.

x = 45.3 [m]

Explanation:

The first thing we should do is a complete description of the problem, doing a Google search, we see a complete description of this problem as well as the question for that problem.

"A small object moves along the x-axis with acceleration a x(t)=−(0.0320m/s3)(15.0s−t). At t=0 the object is at x=-14.0 m and has velocity v0x=7.8m/s. What is the x-coordinate of the object when t=10.0 s?"

Once the problem is identified, we can detail the initial data:

ax(t) = −(0.0320m/s3)(15.0s−t) [m/s^2]

at time t = 0

x = -14 [m]

v0x = 7.8 [m/s]

From the differential calculus, we know that the acceleration is equal to the derivative of the velocity with respect to the time.

After finding the expression of velocity, we know that by the differential calculus that velocity is the derivative of space with respect to time.

x = 45.3 [m]

In the attached document (pdf) we can see the complete solution of the problem.

i dont know but

explanation:

whoever your teacher is, deserves a raise for using pokemon as an example

a and b sometimes have a positive value