AP cal, graphing calculator req.

Question

A particle moves along the x-axis so that the velocity at time t is given by v(t) = t^6-13t^4+12/10t^3+3. At time t = 0, the initial position of the particle is x = 7.

(a) Find the acceleration of the particle at time t = 5

(b) Find ALL values of t in the interval 0≤ t ≤2 for which the speed of the particle is 1.

(c) Find the position of the particle at time t = 4. Is the particle moving toward the origin of away from the origin at time t = 4? justify your answer

(d) During the time interval 0 ≤ t ≤ 4, does the particle return to its initial position? Give a reason for your answer.

Answer 1

v(t) = t⁶ - 13 t⁴ + (12/10) t³ + 3 = t⁶ - 13 t⁴ + (6/5) t³ + 3

(a) Acceleration is the derivative of velocity.

Thus, the acceleration of this particle at time t is given by:

a(t) = v'(t) = 6 t⁵ - 13 * 4 t³ + (6/5) * 3 t² = 6 t⁵ - 52 t³ + (18/5) t²

The acceleration of the particle at time t = 5 is given by:

a(5) = 6 * 5⁵ - 52 * 5³ + (18/5) * 5² = 12,340

(b) This question would require the use of a functionality of the graphing calculator.

An additional consideration is the usage of the term "speed" here whereas initially the term "velocity" was used. If these terms are used properly, the speed would be 1 when the velocity is 1 or when the velocity is -1.

You would graph the function v and look for all the values of t between 0 and 2 for which the function is equal to -1 or equal to 1.

(c) Position is the antiderivative of velocity.

Since we will be taking an antiderivative in a moment, let us recall the power rule for antiderivatives:

For all n ≠ -1,

∫ xⁿ dx = 1/(n + 1) * x^{n + 1} + C

What happens when n = -1?

We can see that it causes a big trouble in the denominator of the formula. Additionally,

∫ x^-1 dx = ∫ (1/x) dx = ln |x| + C

Thus, the position of this particle at time t is given by:

x(t) = ∫v(t) dt

= ∫ [t⁶ - 13 t⁴ + (12/10) t³ + 3]dt

= (1/7) t⁷ - 13 * (1/5) t⁵ + (12/10) * (1/4) t⁴ + 3t + C

= (1/7) t⁷ - (13/5) t⁵ + (3/10) t⁴ + 3t + C

"At time t = 0, the initial position of the particle is x = 7." so x(0) = 7

x(0) = 7

⇒ (1/7) * 0⁷ - 13/5 * 0⁵ + 3/10 * 0⁴ + 3 * 0 + C = 7

⇒ C = 7

x(t) = (1/7) t⁷ - (13/5) t⁵ + (3/10) t⁴ + 3t + 7

Thus, the position of the particle at time t = 4 is given by:

x(4) = (1/7) * 4⁷ - (13/5) * 4⁵ + (3/10) * 4⁴ + 3 * 4 + 7 ≈ -226.03

The velocity of the particle at time t = 4 is given by:

v(4) = 4⁶ - 13 * 4⁴ + (6/5) * 4³ + 3 = 847.8

At time t = 4, at a "negative position" (to the left of the origin), the particle has a positive velocity (so moving towards the right), so it is going towards the origin.

(d) Yes it does. I am not sure what kind of reason is expected / accepted.

Hopefully, stating that "it can be seen on the graph" is good enough :-)