First, we integrate  with respect to  to get velocity:

We know that the particle is not moving at , so

.

Solving this gives us

.

Now, we have to solve for position. To do this, we integrate  with respect to . Thus, we get

.

We know that the particle starts at the origin, so we have to solve:

This implies that , so we have:

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As a special case:

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We begin by integrating the acceleration function and using the initial condition to find the value of the constant of integration:

Now that we have the velocity function we repeat the process to find the position function:

By definition, a given vector  has a perpendicular vector . Given a vector , we therefore know its perpendicular vector is .

We can further verify this by noting that the product of a vector and its perpendicular vector is .

Since , the two vectors are perpendicular to each other. 

By definition, a given vector  has a perpendicular vector . Given a vector , we therefore know its perpendicular vector is .

We can further verify this by noting that the product of a vector and its perpendicular vector is .

Since , the two vectors are perpendicular to each other. 

By definition, a given vector  has a perpendicular vector . Given a vector , we therefore know its perpendicular vector is .

We can further verify this by noting that the product of a vector and its perpendicular vector is .

Since , the two vectors are perpendicular to each other. 

By definition, a given vector  has a perpendicular vector . Given a vector , its perpendicular vector will be . We can further verify this result by noting that the product of two perpendicular vectors is ; since , we know the two vectors are perpendicular to each other. 

In this question, we need to find the minimum of , so we need to take a derivative. But wait! We were given , so we're already one step ahead of the game because  is the derivative of . 

The initial condition  is actually superfluous information because we do not need to integrate.

Setting  we get

Next we need to set each part equal to zero and solving for x.

Therefore,  and  can be disregarded because the problem asks for .

To check if  is a minimum, we can use the second derivative test. The derivative of  is , which is positive at , so  is a minimum.

Mark will be furthest from his home when he is at the school, so we need to maximize this function. We must therefore take the derivative and set it equal to zero.

Use the rule,

 to find the derivative.

This result in,

.

Using the second derivative test, we can show it's a max because the second derivative is  (which is negative at , thus  is a maximum).

We can substitute  back into the  to find where his school is. 

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So his school is at . No distance units were mentioned in the problem, so leaving the answer unitless is acceptable.

To solve this problem, we first have to understand that the derivative of position with respect to time is velocity and the derivative of velocity with respect to time is acceleration. By understanding that concept, we then are able to perform a double intergration to find the position function for the fruit. The general formula for integration is  and if integrating a constant , where C can be any real number. 

The first integration of the acceleration yields 

, however the question states the intial velocity of the fruit is 0, therefore the final velocity equation is .  

Performing the second integration yields 

; we are given the initial position of the fruit to be 20 meters, therefore the final position equation becomes .  

We are asked to find the position of the fruit at , so by plugging in  into our position equation, we find that the position of the fruit at  is 0 meters.

For this problem, it is important to understand that velocity is the derivative of position with respect to time and acceleration is the derivative of velocity with respect to time. Given the initial acceleration equation, we can take the double integral to obtain the position equation. The general formula for integration is 

 and if integrating a constant , where C can be any real number.  

Therefore the integral of the acceleration gives 

, however we can solve for C given that the inital velocity is .  

This makes the velocity euqation .  Taking the integral of the velocity equation, we obtain the position equation .  We are given the position of the ball at 2 seconds to be 70 meters.  Plugging those two values in will allow us to solve for .  

Therefore the final position equation is 

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