Calculus AB Position Velocity & Acceleration

Position Velocity & Acceleration

An object's position is given by x(t) = 2 x³ + 4 x² + x + 20. Find the difference in position of t = 0 and t = 2

• Difference in position is equal to x(2) − x(0)
• x(2) = 2 (2³) + 4 (2²) + 2 + 20 = 16 + 16 + 2 + 20 = 54
• x(0) = 2(0) + 4(0) + 0 + 20 = 20

x(2) − x(0) = 34

An object's position at t ≥ 0 is given by x(t) = −9.8 t² + 20t. Find the value of t when the velocity is zero.

• The velocity can be seen as the derivative of the position function, so we must find when the derivative of x is zero.
• x′(t) = −19.6t + 20
• −19.6t + 20 = 0
• −19.6t = −20

t = [20/19.6]

An object's position with respect to t is given by x(t) = 2(sin5t). Find the velocity and acceleration of the object with respect to t

• x′(t) = v(t) = [d/dt] 2(sin5t)
• v(t) = 2 (cos5t) [d/dt] 5t
• v(t) = 10(cos5t)
• x"(t) = v′(t) = a(t) = [d/dt] 10(cos5t)
• a(t) = 10 (− sin5t) 5 = −50 sin5t

v(t) = 10 cos5t and a(t) = −50 sin5t

A falling ball's position is given by x(t) = 9.8 [m/(s²)] t². How long would it take for the ball to fall 50 meters?

• Solve for t such that x(t) = 50 m
• 9.8 [m/(s²)] t² = 50 m
• t² = [50 m/(9.8 [m/(s²)])]
• t² = [(50 s²)/9.8]
• t = ±√{[(50 s²)/9.8]}
• Since we're talking about a length of time, we're only really interested in the positive value.

t = √{[50/9.8]} s

In the previous problem, what is the ball's speed when it has fallen exactly 50 meters?

• We already know the value of t when the ball has fallen 50 meters. We now need to find the speed at that time.
• v(t) = x′(t) = 9.8 [m/(s²)] (2t)
• v(√{[50/9.8]} s) = 19.6 [m/(s²)] √{[50/9.8]} s

v = 19.6 √{[50/9.8]} [m/s]

A ball is thrown up into the air. Its position is given by x(t) = (−9.8 [m/(s²)]) t² + (20 [m/s]) t. What is the maximum height the ball reaches?

• Let's graph x with respect to t
• 
• From inspection, it looks like it hits a maximum of about 10 m at about the 1 second mark.
• A parabola like this will always hits its maximum (or minimum) when the slope or velocity is equal to zero.
• v(t) = [d/dt] x(t)
• v(t) = [d/dt] (−9.8 [m/(s²)]) t² + [d/dt] (20 [m/s]) t
• v(t) = −19.6 [m/(s²)] t + 20 [m/s]
• Set to zero and solve for t
• −19.6 [m/(s²)] t + 20 [m/s] = 0
• −19.6 [m/(s²)] t = −20 [m/s]
• t = [20/19.6] s
• Plug that time in to the position function
• x([20/19.6] s) = (−9.8 [m/(s²)]) ([20/19.6] s)² + (20 [m/s]) [20/19.6] s

The ball reaches a height of about 10.2  m

An object's position with respect to t is given by x(t) = −9.8 t² + 20t − 30. Find the velocity when t = 0

• v(t) = [d/dt] −9.8 t² + 20t − 30
• v(t) = −19.6 t + 20
• v(0) = −19.6 (0) + 20

v(0) = 20

A solar sail has an estimated acceleration of 13 [km/(hour²)]. Starting from a complete stop, the position is given at t ≥ 0 by x(t) = 13 [km/(hour²)] t². Find the velocity after 20 days.

• v(t) = 13 [km/(hour²)] (2t)
• v(t) = 26 [km/(hour²)] t
• The function is currently in terms of hours and the units should cancel. We can either convert the function to take time in terms of days, or convert the number of days into hours.
• v(24 [hours/day] * 20  days) = 26 [km/(hour²)] (480  hours)

After 20 days, the ship is traveling 12480 [km/hour]

Space probe Voyager 1's position relative to the Sun can be approximately described by x(t) = 61530 [km/hour] t + 1.8 * 10¹⁰ km. Using values from the previous problem, how long would it take that solar sail to achieve greater speed than Voyager 1?

• We need to find when their speeds would be equal. We'll call Voyager 1 ship A, and the solar sail ship B.
• v_A(t) = [d/dt] (61530 [km/hour] t + 1.8 * 10¹⁰ km)
• v_A(t) = 61530 [km/hour]
• From the previous problem we know that v_B(t) = 26 [km/(hour²)] t
• Now we set the two equal and solve for t
• 26 [km/(hour²)] t = 61530 [km/hour]
• t = [61530/26] [km/hour] [(hour²)/km]
• t ≈ 2366.5  hours

It would take approximately 2367  hours or 99  days for the solar sail to achieve greater speed.

An object's position at t ≥ 0 is given by x(t) = −9.8 t² + 20t. Graph an overlay of the position, velocity, and acceleration of the object with t as the horizontal axis.

• v(t) = −9.8 (2t) + 20 = −19.6t + 20
• a(t) = −19.6

Position Velocity & Acceleration

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

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• Intro 0:00
• Position, Velocity, and Acceleration 0:10
• Position Function
• Velocity Function
• Acceleration Function
• Example 1 1:20
• Example 2 6:31
• Example Continue: Velocity When Acceleration is Zero 6:32
• Example 3: Where Is Particle Changing Directions? 8:16
• Example 4: Total Distance Traveled From 0 to 2 Second 11:09
• Example 5: Ball Drop Problem 16:40