V(t) models the velocity of a spaceship.  Find the spaceship's acceleration after 3 seconds.

Remember how position/velocity/acceleration are all related: Velocity is the first derivative of position, and acceleration is the derivative of velocity. 

So, we need to take the derivative of V(t). To find that, we will decrease each exponent by 1 and multiply by the original number:

Thus, we get:

We have one more step, however. We need to find the acceleration after 3 seconds.

So our answer is 198.

Acceleration is the second derivative of position, so the position function must be differentiated twice. 

The first derivative of position, the velocity function, is:

Then the second derivative gives us acceleration:

The particle's accelertation is given by the second derivative of the position: 

By plugging in :

Begin by remembering that the acceleration of a particle can be found by solving for the second derivative of the position function for that particle.  Therefore, for our data, we know:

Then, 

This means that at any time , the acceleration is .

We are looking to apply a given deceleration to a ship so that its velocity is .  Now, since we know that the deceleration is a constant, we can say:

Thus, we know that we can rewrite the velocity equation (for the period of time under consideration, at least) as being:

Thus, we know:

However, since the initial velocity is , we know that  must be .  Therefore, we can write:

Now, we want to know what  must be in order for  to be  at time .  Thus, you can solve the equation:

This means that:

We could say that this is a deceleration by:

(Note, that this also can be solved without integrals, by noting that acceleration is merely the change in velocity over time.)

The instantaneous acceleration is simply the second derivative of the position function with respect to time:

Therefore

To find the instantaneous acceleration at t=3 seconds:

Recall that the accleration of a particle is found by taking the second derivative of the position funciton.  Therefore, we need to do that first.

To do that, you will need to use the chain rule several times on the sine function.  First, take out the  that is the power:

Next, differentiate the :

Thus, we know:

Slightly simplified, this is:

Now, as you can tell, the second derivative will be quite involved.  Therefore it would be easiest to find the acceleration merely by using the first derivative, solving for the change in velocity divided by time.  (This is exactly what the acceleration is.)  The first derivative represents the velocity.  Thus, we need to find:

This is because 

Now, solving for , you get , for  is also .  Therefore, you know that you have:

The position of a particle is defined by the function:

, where  is the number of seconds after the start of the day.

What is the acceleration of the particle from  to ?

Before beginning, there are two things we should recall:

1. The velocity of a body can be found by using the first derivative of its position function.
2. The acceleration of a body can be found by taking the change of velocity over the length of the time period of the change.

Based on this, we can compute the acceleration of the body as being the following:

Therefore, first compute , namely the first derivative of :

Using the product rule, we know that the first term will be:

 or 

Therefore, 

So, to find the acceleration, we need to compute 

Thus, we have:

 or approximately 

Acceleration is given by the time derivative of velocity.

In this case we will use the rules that the derivative of 

 is  

and the derivative of 

 is .

Applying this knowledge we can find the acceleration function to be the following.

Acceleration is given as the second time derivative of position. In this case we will use the rule of exponentials and trigonometrics to solve. Specifically we will use the following rules,

,

,

.

Begin by taking the first derivative:

Afterwards, acceleration may be found by deriving once more: