The coordinates of the midpoint are simply the averages of the coordinates of the two points .

([-1+3]/2, [0+4]/2, [5+1]/2)

=(2/2, 4/2, 6/2)

=(1,2,3)

To solve for this displacement, we can take the double integral of a(t). For simplicity, let us define things as follows:

v(t) = ∫(a(t) dt) and s(t) = ∫(v(t) dt)

The integrals are rather simple:

v(t) = sin(t) + 6t + c₁

To solve for c₁, we know that at t = 0, the initial velocity was 4. This means v(0) = sin(0) + 6 * 0 + c₁ = 4; sin(0) = 0, so we know c₁ = 4

Therefore, v(t) = sin(t) + 6t + 4

s(t) = –cos(t) + 3t² + 4 * t + c₂  (The final constant will drop out when we solve.)

Since we know that the body's position is always increasing (based on the initial positive velocity and an acceleration that must always be positive based on its equation), the total displacement is merely found by subtracting s(2) from s(4):

s(4) = –cos(4) + 3*4² + 16 + c₂ = –cos(4) + 48 + 16 + c₂ = –cos(4) + 64 + c₂

s(2) = –cos(2) + 3*2² + 8 + c₂ = –cos(2) + 12 + 8 + c₂ = –cos(2) + 20 + c₂

s(4) - s(2) = –cos(4) + 64 + cos(2) – 20 = 44.237496784317 (approx.) or 44.24

To find the change in position, we must take the definite integral of the velocity function

∫[4, 8](4t – 3)dt = [4, 8](2t² – 3t) = (2 * 8² – 3 * 8) – (2 * 4² – 3 * 4)

= 2 * 64 – 24 – (2 * 16 – 12) = 104 – 20 = 84

To find the distance traveled by the object over a certain amount of time, we need an equation for its position. We are given an equation for its velocity, so if we integrate that equation from t=1 to t=2 seconds we'll obtain the distance traveled by the object over that interval:

In order to find the distance traveled by an object we need an equation for position. If velocity is the derivative of position, then we must integrate the given equation from t=2 to t=5 to find the total distance traveled by the object over that interval:

To find the midpoint between the two points, the average of the coordinates can be taken:

Where  and 

Plugging in these values we find our midpoint.

To find the displacement of the object, we can integrate the equation for acceleration twice. 

Integrating the acceleration equation once will give:

Because the initial velocity = 2, we can set . Therefore,

Therefore, .

We can now use this in the velocity equation to give

integrating the velocity equation from  to  will give the distance

To find the distance travelled, we must integrate the velocity equation

To find the distance between two points we need to find the difference of the x-coordinates and y-coordinates between the two points.

So the x-coordinate is

The y-coordinate is

So the x-coordinate and y-coordinate is

The distance can be calculated by

So plugging in

We will use the distance formula to show this result. Recall that having two points with coordinates , then the distance between this two points is given by:

.  We will call the first fixed point . To find the second point, note that the equation of the circle with radius 1 is given by:

.  Since we are looking for the upper half, we have by solving for y:

( We need only the upper half in this case , that is why y is .

Hence our second point is .

Using the distance formula we have:. 

We need to simplify this expression a bit:

this gives finally after cancelling and adding: