The acceleration of an object (in m/s^2) is given by the function a(t) = 9 sin(t). the initial velocity of the object is v(0) = -11 m/s. a) find an equation v(t) for the object velocity v(t) =) find the object's displacement (in meters) from time 0 to time 3 meters c) find the total distance traveled by the object from time 0 to time 3 meters

Using integrals, it is found that:

a)

b) The displacement was of -34.27 meters.

c) The total distance traveled was of 34.27 meters.

The acceleration is given by:

Item a:

The velocity is the integrative of the acceleration, hence:

In which K is the integration constant, which is the initial velocity , hence:

Item b:

The position is the integrative of the velocity, hence:

Since the initial position is 0, K = 0, and:

The displacement is:

Hence:

Then

The displacement was of -34.27 meters.

Item c:

The total distance traveled is the absolute value of the integral.

Since it has only one interval, from 0 to 3, it is of 34.27 meters.

A similar problem is given at link

a) Acceleration is the derivative of velocity. By the fundamental theorem of calculus,

so that

b) We get the displacement by integrating the velocity function like above. Assume the object starts at the origin, so that its initial position is . Then its displacement over the time interval [0, 3] is

c) The total distance traveled is the integral of the absolute value of the velocity function:

for and for , so we split the integral into two as

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it