Connecting Position, Velocity, and Acceleration - AP Calculus BC
Card 1 of 30
If $v(t) = 3t - 4$, find $a(t)$.
If $v(t) = 3t - 4$, find $a(t)$.
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$a(t) = 3$. Differentiate the velocity function.
$a(t) = 3$. Differentiate the velocity function.
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If $s(t) = \frac{1}{3}t^3$, what is $v(t)$?
If $s(t) = \frac{1}{3}t^3$, what is $v(t)$?
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$v(t) = t^2$. Differentiate the position function.
$v(t) = t^2$. Differentiate the position function.
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State the formula to find average velocity over $[a, b]$.
State the formula to find average velocity over $[a, b]$.
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$\frac{s(b) - s(a)}{b - a}$. Average rate of change of position over the interval.
$\frac{s(b) - s(a)}{b - a}$. Average rate of change of position over the interval.
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Find $a(3)$ if $v(t) = t^2 - 2t + 1$.
Find $a(3)$ if $v(t) = t^2 - 2t + 1$.
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$a(3) = 4$. Differentiate velocity: $a(t) = 2t - 2$, then evaluate at $t = 3$.
$a(3) = 4$. Differentiate velocity: $a(t) = 2t - 2$, then evaluate at $t = 3$.
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What is the velocity function $v(t)$ if $a(t) = 0$?
What is the velocity function $v(t)$ if $a(t) = 0$?
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$v(t) = C$, a constant. Zero acceleration implies velocity remains constant.
$v(t) = C$, a constant. Zero acceleration implies velocity remains constant.
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Identify the type of motion when $a(t) = 0$.
Identify the type of motion when $a(t) = 0$.
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Uniform motion. Zero acceleration means constant velocity motion.
Uniform motion. Zero acceleration means constant velocity motion.
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Evaluate $v(1)$ if $s(t) = t^3 + 2t$.
Evaluate $v(1)$ if $s(t) = t^3 + 2t$.
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$v(1) = 5$. Differentiate position: $v(t) = 3t^2 + 2$, then evaluate at $t = 1$.
$v(1) = 5$. Differentiate position: $v(t) = 3t^2 + 2$, then evaluate at $t = 1$.
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If $v(t) = 5t - 3$, find $a(t)$.
If $v(t) = 5t - 3$, find $a(t)$.
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$a(t) = 5$. Differentiate the velocity function.
$a(t) = 5$. Differentiate the velocity function.
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What does a negative velocity indicate?
What does a negative velocity indicate?
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Motion in the opposite direction. Negative velocity means moving in the negative direction.
Motion in the opposite direction. Negative velocity means moving in the negative direction.
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If $v(t) = 3t - 4$, find $a(t)$.
If $v(t) = 3t - 4$, find $a(t)$.
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$a(t) = 3$. Differentiate the velocity function.
$a(t) = 3$. Differentiate the velocity function.
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What is the expression for instantaneous velocity?
What is the expression for instantaneous velocity?
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$v(t) = \frac{ds}{dt}$. The derivative of position gives instantaneous velocity.
$v(t) = \frac{ds}{dt}$. The derivative of position gives instantaneous velocity.
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What is $a(t)$ if $v(t)$ is constant?
What is $a(t)$ if $v(t)$ is constant?
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$a(t) = 0$. Derivative of a constant velocity is zero.
$a(t) = 0$. Derivative of a constant velocity is zero.
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Identify the derivative that gives the acceleration function.
Identify the derivative that gives the acceleration function.
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Derivative of the velocity function. Acceleration is found by differentiating velocity.
Derivative of the velocity function. Acceleration is found by differentiating velocity.
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Describe the velocity if acceleration is zero.
Describe the velocity if acceleration is zero.
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Velocity is constant. No acceleration means velocity doesn't change.
Velocity is constant. No acceleration means velocity doesn't change.
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Identify the unit for velocity if position is in meters.
Identify the unit for velocity if position is in meters.
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Meters per second (m/s). Distance per time unit matches the position unit.
Meters per second (m/s). Distance per time unit matches the position unit.
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If $s(t) = t^3 - 3t^2 + t$, find $v(t)$.
If $s(t) = t^3 - 3t^2 + t$, find $v(t)$.
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$v(t) = 3t^2 - 6t + 1$. Differentiate the position function to get velocity.
$v(t) = 3t^2 - 6t + 1$. Differentiate the position function to get velocity.
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What is the physical interpretation of acceleration?
What is the physical interpretation of acceleration?
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Rate of change of velocity with respect to time. Describes how velocity changes with time.
Rate of change of velocity with respect to time. Describes how velocity changes with time.
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What is the physical interpretation of velocity?
What is the physical interpretation of velocity?
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Rate of change of position with respect to time. Describes how position changes with time.
Rate of change of position with respect to time. Describes how position changes with time.
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Calculate the acceleration at $t = 1$ if $v(t) = t^2 + 2t$.
Calculate the acceleration at $t = 1$ if $v(t) = t^2 + 2t$.
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$a(1) = 4$. Differentiate velocity: $a(t) = 2t + 2$, then evaluate at $t = 1$.
$a(1) = 4$. Differentiate velocity: $a(t) = 2t + 2$, then evaluate at $t = 1$.
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What is the initial velocity if $v(t) = 2t + 3$?
What is the initial velocity if $v(t) = 2t + 3$?
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$v(0) = 3$. Substitute $t = 0$ into the velocity function.
$v(0) = 3$. Substitute $t = 0$ into the velocity function.
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If $s(t) = 7t - t^2$, what is $v(t)$?
If $s(t) = 7t - t^2$, what is $v(t)$?
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$v(t) = 7 - 2t$. Differentiate the position function.
$v(t) = 7 - 2t$. Differentiate the position function.
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Find the velocity at $t = 3$ if $s(t) = 4t^2 + 2t + 1$.
Find the velocity at $t = 3$ if $s(t) = 4t^2 + 2t + 1$.
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$v(3) = 26$. Take the derivative: $v(t) = 8t + 2$, then substitute $t = 3$.
$v(3) = 26$. Take the derivative: $v(t) = 8t + 2$, then substitute $t = 3$.
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Evaluate $a(0)$ if $v(t) = 4t^2 - 2t$.
Evaluate $a(0)$ if $v(t) = 4t^2 - 2t$.
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$a(0) = -2$. Differentiate velocity: $a(t) = 8t - 2$, then evaluate at $t = 0$.
$a(0) = -2$. Differentiate velocity: $a(t) = 8t - 2$, then evaluate at $t = 0$.
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Find the velocity at $t = 0$ if $v(t) = t^2 - 4t + 3$.
Find the velocity at $t = 0$ if $v(t) = t^2 - 4t + 3$.
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$v(0) = 3$. Evaluate the velocity function at $t = 0$.
$v(0) = 3$. Evaluate the velocity function at $t = 0$.
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If $v(t) = 2t^2 - 4t + 1$, find $a(t)$.
If $v(t) = 2t^2 - 4t + 1$, find $a(t)$.
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$a(t) = 4t - 4$. Differentiate the velocity function to get acceleration.
$a(t) = 4t - 4$. Differentiate the velocity function to get acceleration.
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What is the formula for acceleration given velocity $v(t)$?
What is the formula for acceleration given velocity $v(t)$?
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Acceleration $a(t) = \frac{dv}{dt}$. Acceleration is the derivative of velocity with respect to time.
Acceleration $a(t) = \frac{dv}{dt}$. Acceleration is the derivative of velocity with respect to time.
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Calculate the average velocity if $s(4) = 10$ and $s(1) = 4$.
Calculate the average velocity if $s(4) = 10$ and $s(1) = 4$.
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$\frac{10 - 4}{4 - 1} = 2$. Apply the average velocity formula with given values.
$\frac{10 - 4}{4 - 1} = 2$. Apply the average velocity formula with given values.
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What is the velocity function $v(t)$ if $a(t) = 6t$?
What is the velocity function $v(t)$ if $a(t) = 6t$?
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$v(t) = 3t^2 + C$. Integrate acceleration to get velocity, adding constant of integration.
$v(t) = 3t^2 + C$. Integrate acceleration to get velocity, adding constant of integration.
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If $v(t) = t^2 + 5$, find $a(t)$.
If $v(t) = t^2 + 5$, find $a(t)$.
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$a(t) = 2t$. Differentiate the velocity function.
$a(t) = 2t$. Differentiate the velocity function.
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Find $v(2)$ if $s(t) = 5t^2 - t$.
Find $v(2)$ if $s(t) = 5t^2 - t$.
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$v(2) = 19$. Differentiate position: $v(t) = 10t - 1$, then evaluate at $t = 2$.
$v(2) = 19$. Differentiate position: $v(t) = 10t - 1$, then evaluate at $t = 2$.
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