This problem can be done using the fundamental theorem of calculus. We hvae

where  is the position at time . So we have

so we need to solve for :

So then the position at  is

We can use the fundamental theorem of calculus for this problem. We have

so we have

This means we just need to solve :

So then the position of the particle at  is 

Begin by finding the velocity function, the integral of acceleration with respect to time:

To find the constant of integration, the fact that the car has a velocity of  when it begins to accelerate:

To find the distance travel, integrate the velocity function with respect to the lower and upper bounds, the start and end times respectively:

V(t) models the velocity of a spaceship. Find a function to model the spaceship's position as a function of time  if P(t) contains the point (2,15).

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

So to get to position, we need to integrate velocity.

To integrate polynomials, we increase the exponent by 1, and then divide  by the new exponent.

Thus, our function so far looks like:

However, we know that it must pass through the point (2,15), so we have to plug the point in and solve for C

So our answer is:

V(t) models the velocity of a spaceship. Find the position of the spaceship after 5 seconds if P(t) (the position function) contains the point (2,15).

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

So to get to position, we need to integrate velocity.

To integrate polynomials, we increase the exponent by 1, and then divide  by the new exponent.

Thus, our function so far looks like:

However, we know that it must pass through the point (2,15), so we have to plug the point in and solve for C

So our function is:

Then, to find it after 5 seconds:

To approach this problem, bear in mind that there are two states for Harold; one in which he's accelerating, and another in which his speed is held constant. Begin with calculating how far he travels from the stone while accelerating:

Since he begins at rest

And make note of his velocity at the five second mark:

Now, to find where he travels while accelerating, integrate once more:

We are told he's initially  feet from a stone:

This is his position after he travels while accelerating. Now to find how far he travels for the one second while his speed is constant, treat this position as a new initial condition, and recall his velocity found above after he finishes accelerating:

The reason the value  is used instead of  is because he only travels for one second in the constant velocity condition.

A position is function is given by the integral of the velocity function with respect to time:

The constant of integration can be found by utilizing the initial position:

And so

Position is given by the integral of the velocity function with respect to time:

To find the constant of integration, use the initial condition:

This yields the function:

Therefore

Using calculus, you know that you can compute this by "working backword" from the acceleration function.  We know that:

.  Since our initial velocity is , it is:

Now, from this, we can compute the position of the particle by computing:

Since we know that the initial height of the ball was , we likewise know:

Thus, at , we have:

Given the acceleration, the position is the second integral of the function. To find the velocity function, the first integral, find the antiderivative of the acceleration function.

This gives

 where C is a constant that can be found using other information from the problem.

We know that at  the velocity is 0 m/s.

Plugging that in we get 

 and we get .  

Therefore, 

.

To find the position function, take the antiderivative again.

 where D is another constant. Using a similar strategy to find the value of the constant we use that at  the position is  m.  and .  

Finally, we get 

.