How to find velocity - CLEP Calculus

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Question

The position of a particle is represented by f(t) = 4t3 – 3t + 15

What is its instantaneous velocity at time t = 3?

Answer

First we must find the simple derivative:

f'(t) = 12t2 – 3

The instantaneous velocity is f'(3) or 12 * 3 * 3 – 3 = 105

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Question

What is the instantaneous velocity at time t = π/2 of a particle whose positional equation is represented by s(t) = 12tan(t/2 + π)?

Answer

The instantaneous velocity is represented by the first derivative of the positional equation.

v(t) = s'(t) = 12 * (1/2) sec2(t/2 + π) = 6sec2(t/2 + π) = 6/cos2(t/2 + π) = 6/((–1)2 cos2(t/2)) = 6/cos2(t/2)

Based on the nature of the cosine, we know that 6/cos2(t/2 + π) = 6/((–1)2 cos2(t/2)) = 6/cos2(t/2)

To solve, v(π/2) = 6/cos2(t/2) = 6/((1/√2)2) = 6/(1/2) = 12

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Question

What is the instantaneous velocity at time t=π/2 of a particle whose positional equation is represented by s(t) = 12cos2(t/2 + π)?

Answer

The instantaneous velocity is represented by the first derivative of the positional equation. This is found by using the chain rule both on the square of the cosine function and the function itself.

v(t) = s'(t) = 12 * 2 cos(t/2 + π) * (–sin(t/2 + π)) * (1/2) = –12cos(t/2 + π)sin(t/2 + π)

Given what we know about the cosine and sine functions, we know cos(t/2 + π) = –cos(t/2) and sin(t/2 + π) = –sin(t/2)

Therefore, v(t) = –12(–sin(t/2))(–cos(t/2)) = –12sin(t/2)cos(t/2)

v(π/2) = –12sin(π/4)cos(π/4) = –12(1/√2)(1/√2) = –12 * (1/2) = –6

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Question

The acceleration of a particle is given by the function . What is the particle's average velocity over the interval to , if it has an initial velocity of zero?

Answer

Velocity can be found by integrating acceleration with respect to time:

$v(t)=\int a(t)dt=\int (6t^2+4t^3)dt$

To find the integration of constant, utilize the initial velocity:

Now, to find the average velocity, integrate the velocity function once more over the interval and divide by the interval:

$v_{avg}=\frac{1}{4-2}\int_{2}^{4}(2t^3+t^4)dt=\frac{1}{4-2}(\frac{t^4}{2}+\frac{t^5}{5}|_2^4)$

$v_{avg}=\frac{1}{2}((\frac{256}{2}+\frac{1024}{5})-(\frac{16}{2}+\frac{32}{5}))$

$v_{avg}=\frac{1}{2}(\frac{1592}{5})=\frac{796}{5}$

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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

If during the first 15 seconds of its flight the displacement of a spacecraft is given by the equation 21.25x + x2 feet, what is its velocity at 10 seconds, given that its initial velocity is 0?

Answer

To find the velocity function based on displacement, use the first derivative of f(x) = 21.25x + x2. f'(x) = 1.25 * ln(2) * 21.25x + 2x

The velocity at x = 10 would therefore be: f'(10) = 1.25 * log(2) * 21.25 * 10 + 2 * 10 = 1.25 * 212.5 * log(2) + 20 feet/second

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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

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 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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