- The position of a particle moving along the x-axis is given by s(t)=t³-6t²+9t. Sal analyzes it to find the times when the particle is 'speeding up.'
- AP Calculus AB Particle Motion Worksheet In #1-5, answer the following questions for each position function s(t) in meters where t is in seconds if a particle is moving along the x-axis.
- MOTION OF A PARTICLE IN A RESISTING MEDIUM USING FRACTIONAL CALCULUS APPROACH J. Juan ROSALES GARCIA 1, M. GUIA CALDERON 1, Juan MARTINEZ ORTIZ 2, Dumitru BALEANU 3,4,5 1 Universidad de Guanajuato Department of Electrical Engineering, DICIS, Carretera Salamanca-Valle de Santiago, Comunidad de Palo Blanco, Salamanca, Guanajuato, Mexico.
Graphs of Position, Velocity, Speed, and Acceleration for a particle moving on the horizontal line y=3. Its position at time, t, is given by p(t)=2cos(pi/4.t) Move the slider, t, over the interval 0particle's motion and the related graphs. When is the particle speeding up, slowing. AB 2007-2012 Function Max Min, Inflection Points, Tangent Lines OR Solutions.pdf.
2011-2012 Particle Motion Definition and Calculus
The position of a particle (in inches) moving along the x-axis after t seconds have elapsed is given by the following equation:
s = f(t) = t4 – 2t3 – 6t2 + 9t
(a) Calculate the velocity of the particle at time t.
(b) Compute the particle's velocity at t = 1, 2, and 4 seconds.
(c) When is the particle at rest?
(d) When is the particle moving in the forward (positive) direction?
(e) Calculate total distance traveled by the particle (i.e., forwards and backwards) after t = 5 seconds.
(f) Calculate the acceleration of the particle after 4 seconds.
(g) When is the speed of the particle constant?
Solution:
(a) The velocity is the derivative of position, so the velocity is v(t) = 4t3 – 6t2 – 12t + 9.
(b) Simply plug into the velocity equation to get: v(1) = –5 in/sec, v(2) = –7 in/sec, v(4) = 121 in/sec.
(c) If you graph the velocity function on your calculator, you see that it appears to pass through x = –1.5. Use synthetic division to ensure that this is true and to factor the equation. You will get the following:
(t – 1/2)(4t2 – 12t + 6)
Now, use the quadratic formula to solve the quadratic part, and you'll see that the velocity equals zero (in other words, is stopped) when t = –1.5, 0.6339745962, 2.366025404. Even though you can round to the third decimal place, you need to use these values for the remainder of the problem.
(d) If you plug in values into the velocity equation between the x-intercepts above, you will get positive values on the intervals (–1.5, 0.6339) and (2.366, ∞). Note that it doesn't quite make sense to have negative time, so (0, 0.6339) is just as acceptable, and perhaps more so, for the first interval. We do this because positive velocity implies forward movement.
(e) First, substitute the 'turn points' you found in part (c) into the position equation. When the velocity equals zero in this problem, the particle is stopping because it is turning to go the other way. You find thats(0.6339) = 2.946152423, s(2.366) = –7.446152423, and s(5) = 270. Note that the negative x-intercept is ignored because you cannot move back in time.
These numbers represent how far the particle is from the origin at specific times. So, the particle moves 2.9 inches to the right of the origin, then moves 7.44 inches left of it, and finally ends up 270 inches to the right of it. By the time t = 2.366 seconds, the particle has traveled to the right 2.9 inches, back 2.9 inches to the origin, then left 7.4 more inches. It then moves 7.4 inches back to the origin and ends up 270 more inches to the right of it. The final answer is 290.785 inches.
(f) The acceleration is the derivative of velocity, so a(t) = v'(t) = 12t2 – 12t – 12. The acceleration at t = 4 seconds is a(4) = 132 in/sec2.
(g) Set the acceleration equal to zero and solve using the quadratic equation: t = –0.618 sec or 1.618 sec.
The video may take a few seconds to load.Having trouble Viewing Video content? Some browsers do not support this version – Try a different browser.
Here are the resources you will need for the AP Calculus AB free response questions. Be sure to try out the prior year questions and check out the video solutions. Working through these problems is an important part of your FRQ test prep.
2017 & 2018 Free Response
At the College Board website you can access the prior year free response questions. You should definitely review these as part of your FRQ test prep.
Graphs of Position, Velocity, Speed, and Acceleration for a particle moving on the horizontal line y=3. Its position at time, t, is given by p(t)=2cos(pi/4.t) Move the slider, t, over the interval 0particle's motion and the related graphs. When is the particle speeding up, slowing. AB 2007-2012 Function Max Min, Inflection Points, Tangent Lines OR Solutions.pdf.
2011-2012 Particle Motion Definition and Calculus
The position of a particle (in inches) moving along the x-axis after t seconds have elapsed is given by the following equation:
s = f(t) = t4 – 2t3 – 6t2 + 9t
(a) Calculate the velocity of the particle at time t.
(b) Compute the particle's velocity at t = 1, 2, and 4 seconds.
(c) When is the particle at rest?
(d) When is the particle moving in the forward (positive) direction?
(e) Calculate total distance traveled by the particle (i.e., forwards and backwards) after t = 5 seconds.
(f) Calculate the acceleration of the particle after 4 seconds.
(g) When is the speed of the particle constant?
Solution:
(a) The velocity is the derivative of position, so the velocity is v(t) = 4t3 – 6t2 – 12t + 9.
(b) Simply plug into the velocity equation to get: v(1) = –5 in/sec, v(2) = –7 in/sec, v(4) = 121 in/sec.
(c) If you graph the velocity function on your calculator, you see that it appears to pass through x = –1.5. Use synthetic division to ensure that this is true and to factor the equation. You will get the following:
(t – 1/2)(4t2 – 12t + 6)
Now, use the quadratic formula to solve the quadratic part, and you'll see that the velocity equals zero (in other words, is stopped) when t = –1.5, 0.6339745962, 2.366025404. Even though you can round to the third decimal place, you need to use these values for the remainder of the problem.
(d) If you plug in values into the velocity equation between the x-intercepts above, you will get positive values on the intervals (–1.5, 0.6339) and (2.366, ∞). Note that it doesn't quite make sense to have negative time, so (0, 0.6339) is just as acceptable, and perhaps more so, for the first interval. We do this because positive velocity implies forward movement.
(e) First, substitute the 'turn points' you found in part (c) into the position equation. When the velocity equals zero in this problem, the particle is stopping because it is turning to go the other way. You find thats(0.6339) = 2.946152423, s(2.366) = –7.446152423, and s(5) = 270. Note that the negative x-intercept is ignored because you cannot move back in time.
These numbers represent how far the particle is from the origin at specific times. So, the particle moves 2.9 inches to the right of the origin, then moves 7.44 inches left of it, and finally ends up 270 inches to the right of it. By the time t = 2.366 seconds, the particle has traveled to the right 2.9 inches, back 2.9 inches to the origin, then left 7.4 more inches. It then moves 7.4 inches back to the origin and ends up 270 more inches to the right of it. The final answer is 290.785 inches.
(f) The acceleration is the derivative of velocity, so a(t) = v'(t) = 12t2 – 12t – 12. The acceleration at t = 4 seconds is a(4) = 132 in/sec2.
(g) Set the acceleration equal to zero and solve using the quadratic equation: t = –0.618 sec or 1.618 sec.
The video may take a few seconds to load.Having trouble Viewing Video content? Some browsers do not support this version – Try a different browser.
Here are the resources you will need for the AP Calculus AB free response questions. Be sure to try out the prior year questions and check out the video solutions. Working through these problems is an important part of your FRQ test prep.
2017 & 2018 Free Response
At the College Board website you can access the prior year free response questions. You should definitely review these as part of your FRQ test prep.
2018 FRQ Solutions
In this series of videos, Tom Cochran solves the 2018 AP Calculus AB free response questions. Clear solutions that are easy to understand.
2017 FRQ Solutions
More great videos from Mr. Cochran. In these he solves the 2017 AP Calc free response questions. Be sure to try the questions first, then watch the videos.
Sample Questions
The sample question packet from the College Board includes some samples of the updated AP Calculus AB free response questions. Good for test prep and review.
Course Description
Mario Particle Motion Ap Calculus Solver
The official course description includes some free response practice questions as well. The AB FRQ questions start on page 67 of this packet.
Mario Particle Motion Ap Calculus Notes
AP Calculus | Practice Exams | Free Response | Notes | Videos | Study Guides
