# 2.1.314 AP Calculus Exam a particle moves along the x-axis......

• MHB
• karush
In summary, the conversation discusses a common problem in calculus involving particles moving towards the left. It also includes a graph and equations to demonstrate the concept. The total distance and acceleration are also calculated, and it is determined that the particle's speed is increasing at a certain point. Finally, the conversation ends with the mention of an AP calculus exam question.
karush
Gold Member
MHB
View attachment 9347
ok again I used an image since there are macros and image

I know this is a very common problem in calculus but think most still stumble over it
inserted the graph of v(t) and v'(t) and think for v'(t) when the graph is below the x-axis that participle is moving to the left

the integral has a - interval but I think the total is an absolute value...

my take on some of it.

finally this will be my last AP calculus exam question for a while
I was surprized how many views these got... must be a big concern for many

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(a) The particle moves toward the left when $v(t) < 0$

$\cos\left(\dfrac{\pi}{6} t \right) < 0 \text{ in the interval } 3 < t < 9$

(b) total distance $$\displaystyle = \int_0^6 |v(t)| \, dt$$

(c) $a(t) = v'(t) = -\dfrac{\pi}{6} \sin\left(\dfrac{\pi}{6} t \right)$

$a(4) = -\dfrac{\pi}{6} \sin\left(\dfrac{2\pi}{3} \right) < 0$

$v(4) = \cos\left(\dfrac{2\pi}{3} \right) < 0$

Since $a(4)$ and $v(4)$ have the same sign, the particle's speed is increasing

(d) $$\displaystyle x(4) = x(0) + \int_0^4 v(t) \, dt = -2 + \bigg[\dfrac{6}{\pi}\sin\left(\dfrac{\pi}{6} t \right)\bigg]_0^4 = -2+\dfrac{6}{\pi} \bigg[\dfrac{\sqrt{3}}{2} - 0 \bigg]$$

## 1. What is the significance of the "2.1.314" in the title of the AP Calculus Exam?

The "2.1.314" refers to the specific section and problem number in the AP Calculus curriculum. In this case, it is problem 314 in section 2.1, which covers derivatives and their applications.

## 2. How is the particle's position related to its velocity and acceleration?

The particle's position is related to its velocity through the first derivative, and its acceleration through the second derivative. This means that the particle's position can be determined by integrating its velocity function, and its velocity can be determined by differentiating its position function.

## 3. What does it mean for a particle to "move along the x-axis"?

When a particle moves along the x-axis, it means that its position is changing only in the horizontal direction. This is often represented by a graph with time on the x-axis and position on the y-axis.

## 4. How can the particle's position, velocity, and acceleration be represented mathematically?

The particle's position can be represented by a function of time, usually denoted as x(t). Its velocity can be represented by the first derivative of the position function, v(t) = x'(t). And its acceleration can be represented by the second derivative of the position function, a(t) = x''(t).

## 5. What is the purpose of using calculus to analyze a particle's motion?

Calculus allows us to analyze the instantaneous changes in a particle's position, velocity, and acceleration. This is useful for understanding the particle's behavior and predicting its future motion. It also allows us to find maximum and minimum values, which can be helpful in optimization problems.

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