Back to AI Flashcard MakerMathematics /AP Calculus AB: 7.1.1 Acceleration and the Derivative

AP Calculus AB: 7.1.1 Acceleration and the Derivative

Mathematics9 CardsCreated 3 months ago

This set of flashcards explains the relationship between position, velocity, and acceleration through derivatives. It highlights how tangent lines and derivatives help approximate function values, find roots, and optimize outputs. The cards also include practical examples such as calculating acceleration from velocity functions.

Acceleration and the Derivative

  • Velocity is the rate of change of position. Acceleration is the rate of change of velocity.

  • Tangent lines can be used to approximate functions that are difficult to evaluate. The slopes of tangent lines can be used to optimize outputs such as profits and
    areas.

Tap or swipe ↕ to flip
Swipe ←→Navigate
1/9

Key Terms

Term
Definition

Acceleration and the Derivative

  • Velocity is the rate of change of position. Acceleration is the rate of change of velocity.

  • Tangent lines can be used to app...

note

  • Velocity is the rate of change of position.

  • Acceleration is the rate of change of velocity. So
    the velocity function is t...

Suppose an object falling out of an airplane has a velocity of v (t) = −32t + 64 where t is in seconds and v is in feet per second. What is the acceleration of the object when t = 1?

−32 ft/sec2

Given that a particular moving object’s velocity is given by the equation v(t)=−32t+110, what is the equation for the object’s acceleration?

a (t) = −32

Given that a particular moving object’s velocity is given by the equation v(t)=65t−2t^2, what is the equation for the object’s acceleration?

a (t) = 65 − 4t`

Given that a particular moving object’s velocity is given by the equation v(t)=65−3t, what is the equation for the object’s acceleration?

a (t) = −3

Related Flashcard Decks

Study Tips

  • Press F to enter focus mode for distraction-free studying
  • Review cards regularly to improve retention
  • Try to recall the answer before flipping the card
  • Share this deck with friends to study together
TermDefinition

Acceleration and the Derivative

  • Velocity is the rate of change of position. Acceleration is the rate of change of velocity.

  • Tangent lines can be used to approximate functions that are difficult to evaluate. The slopes of tangent lines can be used to optimize outputs such as profits and
    areas.

note

  • Velocity is the rate of change of position.

  • Acceleration is the rate of change of velocity. So
    the velocity function is the derivative of the position
    function and the acceleration function is the
    derivative of the velocity function.

  • The connection between instantaneous rate, the
    derivative, and the slope of the tangent line give rise
    to many different applications of differential calculus.
    Tangent lines can be used to approximate the
    values of hard-to-evaluate functions as well as the
    roots of functions.

  • The derivative can also be used to optimize
    function outputs. The behavior of tangent lines can
    tell you where functions attain maximum and
    minimum values.

Suppose an object falling out of an airplane has a velocity of v (t) = −32t + 64 where t is in seconds and v is in feet per second. What is the acceleration of the object when t = 1?

−32 ft/sec2

Given that a particular moving object’s velocity is given by the equation v(t)=−32t+110, what is the equation for the object’s acceleration?

a (t) = −32

Given that a particular moving object’s velocity is given by the equation v(t)=65t−2t^2, what is the equation for the object’s acceleration?

a (t) = 65 − 4t`

Given that a particular moving object’s velocity is given by the equation v(t)=65−3t, what is the equation for the object’s acceleration?

a (t) = −3

Which of the following statements about acceleration is true?

Acceleration is the rate of change in velocity.

A car is moving at a constant acceleration of 3 m / sec2. If the car is moving with a velocity of 20 m / sec at t = 0, how fast is the car moving when t = 3?

29 m / sec

A jogger is moving at a constant velocity of 8 ft / sec. How far has the jogger moved after 20 seconds?

160 feet