Velocity - AP Calculus AB

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Question

An object's position is described by the given function:

.

Given this information is the object's velocity constant or changing? What is the velocity when ?

Answer

The given equation is used to find position. Based on what we know about derivatives, if we take the derivative of position with respect to time we will be finding the change in position over time, which is velocity. So the first step is to take the derivative. We will get

to represent our velocity. Since our function for velocity is a constant, we know that the velocity will always be 16. To confirm this, we can take the derivative of our velocity to find acceleration. The derivative of a constant is 0, therefore, there is no acceleration or change in velocity, thus velocity is constant at 16.

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Question

A weight hanging from a spring is stretched down 3 units beyond its rest position and released at time t=0 to bob up and down. Its position at any later time t is

What is its velocity at time ?

Answer

$v=\frac{ds}{dt}=\frac{d}{dt}(3cos(t))= -3\sin(t)$

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Question

The position of a particle at time is given by . What is the particle's velocity at time

Answer

The velocity function is given by the derivative of the position function. So here . Plugging 3 in for gives 16.

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Question

The position of a particle is given by $s(t)=\frac{2t^2+1}{t+1}$ . Find the velocity at .

Answer

The velocity is given as the derivative of the position function, or

$v(t)=\frac{d(s(t))}{dt}$ .

We can use the quotient rule to find the derivative of the position function and then evaluate that at . The quotient rule states that

$\left[\frac{f(x)}{g(x)}\right]'=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$ .

In this case, $f(t)=2t^2+1\Rightarrow f'(t)=4t$ and $g(t)=t+1\Rightarrow g'(t)=1$ .

We can now substitute these values in to get

$\left[\frac{f(t)}{g(t)}\right]'=\frac{(t+1)(4t)-(2t^2+1)(1)}{(t+1)^2}=\frac{2t^2+4t-1}{(t+1)^2}$ .

Evalusting this at gives us $\frac{22^2+42-1}{(2+1)^2}=\frac{16-1}{3^2}=\frac{15}{9}$ .

So the answer is $\frac{15}{9}$ .

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Question

Find the velocity function if the position function is given as: .

Answer

There are three terms in this problem that has to be derived. The derivative of the position function, or the velocity function, represents the slope of the position function.

The derivative of can be solved by using the power rule, which is:

$2\cdot3x^\left (2-1\right)$

Therefore. the derivative of is .

The derivative of is by using the constant multiple rule.

The derivative of is since derivatives of constants equal to zero.

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Question

Find the velocity function given the position function: $s(t)= -25t^2 +\frac{2}{t}$ .

Answer

The derivatives can be solved term by term.

First, find the derivative of . This can be done by the power rule.

Find the derivative of $\frac{2}{t}$ . Rewrite this as $2t^\left ( -1)\right \left$ to compute by power rule.

$-2t^\left ( -2)\right \left$

$-\frac{2}{t^2}$

Therefore, the velocity function is: $-50t-\frac{2}{t^2}$

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Question

Find the velocity of a function if the acceleration is: .

Answer

To find the velocity given the acceleration function, we will need to integrate the acceleration function.

$\int a(t) dt = \int (4t+6) dt = 2t^2+6t$

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Question

If models the distance of a projectile as a function of time, find the velocity of the projectile at .

$h(t)=\frac{4x^3}{3}+\frac{4.5x^2}{3}+3x-7$

Answer

We are given a function dealing with distance and asked to find a velocity. recall that velocity is the first derivative of position. Find the first derivative of h(t) and evaluate at t=15.

$h(t)=\frac{4t^3}{3}+\frac{4.5t^2}{3}+3t-7$

$948\frac{m}{s}$

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Question

Find the velocity at if the acceleration function is: .

Answer

The velocity function can be obtained by integrating the acceleration function.

$v(t)=\int a(t)dt= \int 1 dt = t$

Since we are finding the velocity at , substitute this into the velocity function.

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Question

Find the velocity at $t=\pi$ if the position function of a spring is: .

Answer

To find the velocity function, take the derivative of the position function.

$v(t)=\frac{\mathrm{d} }{\mathrm{d} t} s(t) = -5:sin:t$

Substitute $t=\pi$ into the velocity function.

$v(t=\pi)=-5:sin:\pi = 0$

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Question

The position of an object is given by the following equation:

$s(t)=\frac{2}{3}t^3+\frac{5}{2}t^2-13t+27$

Determine the equation for the velocity of the object.

Answer

Velocity is the derivative of position, so in order to find the equation for the velocity of an object, all we must do is take the derivative of the equation for its position:

$s(t)=\frac{2}{3}t^3+\frac{5}{2}t^2-13t+27$

We will use the power rule to get the derivative.

$f(x)=ax^n \rightarrow f'(x)=a \cdot n x^{n-1}$

Therefore we get,

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Question

The acceleration of an object is given by the following equation:

$a(t)=\frac{3}{4}t+\frac{5}{8}$

If $v(0)=1 \ m/s$ , find the velocity of the object at seconds.

Answer

Acceleration is the derivative of velocity, so in order to obtain in equation for the velocity we must integrate the equation for acceleration with respect to time:

$v(t)=\int a(t)dt=\int \left(\frac{3}{4}t+\frac{5}{8}\right)dt=\frac{3}{8}t^2+\frac{5}{8}t+C$

Now we must use the given initial condition, v(0)=1, to solve for C:

$v(t)=\frac{3}{8}t^2+\frac{5}{8}t+C$

$v(0)=0+0+C=1\rightarrow C=1$

$v(t)=\frac{3}{8}t^2+\frac{5}{8}t+1$

Now we can simply plug in t=2 seconds to find the velocity of the object at that time:

$v(2)=\frac{3}{8}(2)^2+\frac{5}{8}(2)+1=\frac{12}{8}+\frac{10}{8}+1=\frac{30}{8}=\frac{15}{4}$

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Question

The displacement of a spacecraft is defined by , what is the instantaneous velocity of the ship at seconds?

Answer

To find the velocity of the of the spacecraft, we can differentiate the position equation:

Now we can use in the velocity equation to find the velocity at 3 seconds.

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Question

The position of a particle is represented by $f(x) = e^{x}+3x^2$ . What is the velocity at ?

Answer

Differentiate the position equation, to get the velocity equation

$f(x) = e^{x}+3x^2$

$v(x) = e^x +3\cdot2 x^{2-1}=e^{x} + 6x$

Now we plug 4 into the equation to find the velocity

$v(4) = e^{(4)}+6(4)$

is approximately equal to 2.72. Therefore

$e^{(4)} + 6(4) = 2.72^{(4)} +24 \approx 55+24= 79$

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Question

What is the instantaneous velocity at $t = \frac{\pi}{2}$ of a particle whose positional equation is represented by $x(t) = 12\cos^2(t)$ ?

Answer

We find the velocity equation by differentiating the position equation. Since cos(t) is also a function we need to use the power rule along with the chain rule. This states to take the derivative of the outside function and multiply it by the derivative of the inside function. In math terms this is as follows:

$v(t) = 12\cdot 2 \cdot\cos(t)(-\sin(t)) = -24\cos(t)\sin(t)$

Using $\frac{\pi}{2}$ as the value of

$v\left(\frac{\pi}{2}\right) = -24\cos\left(\frac{\pi}{2}\right)\sin\left(\frac{\pi}{2}\right) = -24(0)(1) = 0$

Therefore the velocity at $t = \frac{\pi}{2}$ is 0

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Question

A weight hanging from a spring is stretched down by units from its rest position and released at time equals zero to bob up and down. Its position is given by . What is its velocity at time ?

Answer

To find the velocity of the weight, we can differentiate the position equation.

Doing so gives us:

$v(t)=f(t)\cdot g'(t)+g(t)\cdot f'(t)$

where,

$g(t)=\cos(t); g'(t)=-\sin(t)$

thus,

$v(t) = -5\sin(t)$

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Question

The position of a particle at time is given by . What is the particle's velocity at time ?

Answer

To find the velocity, we must first find the velocity equation by differentiating the position equation of the particle.

We can now use 7 as the value for to give

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Question

The position function of a particle is given as: $s(t) = 9t^2+t-\sin(t)$ . What is the velocity function?

Answer

To find the velocity function from the position function, we only need to differentiate the position function.

$v(t)=2\cdot 9 t^{2-1}+1t^{1-1}-\cos(t)$

$v(t) = 18t+1-\cos(t)$

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Question

Find the velocity of a function if the acceleration is: $a(t) = \sin(t) +2t$

Answer

To find the velocity, we integrate the acceleration equation

$v(t) = \int\sin(t)dt + \int2tdt$

$= -\cos(t)+\frac{2t^2}{2}$

$= -\cos(t)+t^2$

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Question

What is the velocity function when the position function is given by

.

Answer

To find the velocity function, we need to find the derivative of the position function.

So lets take the derivative of with respect to .

The derivative of is because of Power Rule:

$f(x)=a^n \rightarrow \ f'(x)=na^{n-1}$

$p(t)=t^2 \rightarrow p'(t)=2t^{2-1}=2t$

The derivative of is due to Power Rule

$p(t)=t \rightarrow p'(t)=1t^{1-1}=1t^0=1(1)=1$

So...

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