The acceleration function is the sencond derivative of the position function, so we must derive the given equation twice.

One of the properties of integrals is ,

so 

.

By the fundemental theorem of Calculus, the derivative of  is , so 

Then, the second derivative, using the power rule,  is .

Acceleration can be found as the time derivative of velocity.

For velocity

The acceleration function is

The initial acceleration is

So to find the time at which the acceleration is half of this, we can use the equation:

Velocity is the derivative of position and acceleration is the derivative of velocity. This means that the second derivative of position is acceleration.

The derivative of  is  according to the power rule.

Applying the power rule once we will find the velocity equation. 

Applying the power rule a second time we will find the acceleration equation.

Acceleration is the second time derivative of position and the first time derivative of acceleration. Begin by finding the velocity functions by taking the derivative with respect to time.

For the position functions

The velocity functions can be found as follows using

The product rule

And the chain rule 

Which in the case of trigonmetric derivatives translates to

Giving the following results:

And the acceleration functions are in turn found using the above rules to be

At time :

Acceleration is found by taking the time derivative of velocity.

For the velocity function

The derivative of functions of the form  is , so the acceleration function can be found to be

To find the time when the scalar values of these two properties are equal, set the equations equal to each other:

The two roots are .

Since a negative time is not possible, the two values are equivalent at time

For the acceleration to be constant, it must be the same for all positive values of time. For this to be true, the time term should be absent from the acceleration equation.

Acceleration is the time derivative of velocity. The only velocity function which, when derived, gives an acceleration function without the time term is

The derivative of functions of the form  is ; when , , i.e. only a constant remains. Furthermore, the derivative of just a constant is always zero.

The acceleration can be found to be

Acceleration is the time derivative of velocity:

However, in this problem, we are told that

Assume that both statements hold true and we can write the equation:

We need an expression for  such that this statement holds true.

Consider known common derivatives which are more oscillatory:

There are the trigonometric functions:

And the exponential function:

Note that in the above expressions, the terms inside of the parentheses/exponent have been kept simple, though they can be made more rigorous.

In particular, consider the form of exponential:

This is very similar to the equation we found earlier, 

For a velocity of the form

, the derivative, i.e. the acceleration would be

Given our initial conditions, we can find n and C:

Now we can consider our acceleration:

Cancelling out like terms yields

.

If the integral shown gives the displacement on that interval, then the integrand must be the velocity function on that interval. Consequently, the acceleration function is given by the derivative of the integrand:

The answer follows directly from the algebra. Recall that

Velocity is the derivative of position and acceleration is the derivative of velocity. That means we have to take the second derivative of position to find the acceleration. The derivative of  is . So

Acceleration of a particle can be found by taking the derivative of the velocity function with respect to time. Recall that a derivative gives the rate of change of some parameter, relative to the change of some other variable. When we take the derivative of velocity with respect to time, we are evaluating how velocity changes over time; i.e acceleration! This is just like finding velocity by taking the time derivative of the position function.

To take the derivative of the function

We'll need to make use of the following derivative rule:

Derivative of an exponential: 

Note that u  may represent large functions, and not just individual variables!

Using the above properties, the acceleration function is