Derivative interpreted as an instantaneous rate of change

Practice Questions

AP Calculus AB › Derivative interpreted as an instantaneous rate of change

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$\begin{align*}&\text{At any given time, a particle has the position :}\&p(t)=cos(4\cdot \pi t) - 2t - sin(2\cdot \pi t) - 3\&\text{Calculate its acceleration at time }t=4\end{align*}$

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\&p(t)=6t - sin(\frac{(5\cdot \pi t)}{2}) + 5\&\text{What is its velocity at time }t=6\text{?}\end{align*}$

$\begin{align*}&\text{A particle at any point in time has the position :}\&p(t)=5t + sin(3\cdot \pi t) - 6t^{2} + 1\&\text{Determine its acceleration at time }t=3\end{align*}$

A particle is traveling in a straight line along the x-axis with position function $x(t) = t+\sin(t)$ . What is the instantaneous rate of change in the particle's position at time $\pi$ seconds?

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\&p(t)=- 3t - cos(5\cdot \pi t) - sin(\frac{(7\cdot \pi t)}{2}) - 5\&\text{What is its velocity at time }t=3\text{?}\end{align*}$

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\&p(t)=cos(\frac{(9\cdot \pi t)}{2}) - 2t + cos(\frac{(17\cdot \pi t)}{2}) - 1\&\text{What is its velocity at time }t=1\text{?}\end{align*}$

The amount of bacteria in a dish (as a function of time) is modeled according to the following expression:

$b(t)=10e^{2t}-2t$

At what time will the amount of bacteria be growing at a rate of 38?

$\begin{align*}&\text{At any given time, a particle has the position :}\&p(t)=3t^{2} - sin(9\cdot \pi t) - 4t + 2\&\text{Calculate its acceleration at time }t=6\end{align*}$