Spatial Calculus - AP Calculus AB

Card 0 of 14378

Question

Suppose the acceleration of a particle is two at . The acceleration of the particle is three when . What is the velocity at if the velocity at is zero?

Answer

By knowing the two accelerations at two different times, it is possible to find the acceleration function. Use the slope and slope-intercept equation to determine the acceleration function. The two points, where acceleration is the dependent variable , are and .

Find the slope.

$m = \frac{y_2-y_1}{x_2-x_1 } = \frac{3-2}{3-1} = \frac{1}{2}$

Find the y-intercept of the acceleration function and choose a point for substitution.

$2=(\frac{1}{2})(1)+b$

$2=\frac{1}{2}+b$

$b=\frac{3}{2}$

The acceleration function is:

$a(t)= \frac{1}{2}t+\frac{3}{2}$

Integrate this function to obtain the velocity.

$v(t)= \int a(t) :dt =\int (\frac{1}{2}t+\frac{3}{2}):dt = \frac{t^2}{4} + \frac{3}{2}t + C$

Substitute the initial condition where velocity is zero when the time is at zero.

$0= \frac{0^2}{4} + \frac{3}{2}(0)+ C$

Write the velocity function.

$v(t) =\frac{t^2}{4} + \frac{3}{2}t$

Substitute into the velocity function.

$v(t) =\frac{3^2}{4} + \frac{3}{2}(3) = \frac{9}{4}+ \frac{9}{2}=\frac{9}{4}+ \frac{18}{4}= \frac{27}{4}$

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Question

Find the velocity of the particle if the acceleration of the particle is given by the following function:

Answer

To find the velocity function from the acceleration function, we must integrate:

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

The integral was performed using the following rule:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$

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Question

Find velocity at given the following position function.

Answer

To find velocity, simply differentiate the position function and then plug in . Thus,

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Question

Find the velocity function if the position function is

Answer

In order to find the velocity function from the position function we must derive the position function

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

When taking the derivative we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}[x^n]=nx^{n-1}$

Applying this rule term by term we get

$\frac{d}{dt}[-30t-16t^2]=-301t^{1-1}-162t^{2-1}$

As such

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Question

Find the acceleration function if the position function is .

Answer

Acceleration is equal to the second derivative of position with respect to time.

Recall the following rules to help solve the problem.

Power Rule:
$\frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}$

Differentiation rules for sine and cosine:

$\\frac{\mathrm{d} }{\mathrm{d} t}(asin(bt)) = abcos(bt)\ \frac{\mathrm{d} }{\mathrm{d} t}(acos(bt)) = -absin(bt)\$

Therefore the acceleration function, by the differentiation rule for cosine and the power rule, is

$\a(t) = \frac{\mathrm{d} }{\mathrm{d} t}\left(\frac{\mathrm{d} }{\mathrm{d} t}(cos(t)+3t^2 - 2)\right)\ \ a(t)= \frac{\mathrm{d} }{\mathrm{d} t}(-sin(t)+6t)\ \ a(t)= 6 - cos(t)$

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Question

Find velocity at given the following position equation.

Answer

To solve, simply differentiate using the power rule and then plug in .

Recall the power rule:

$f'(x)=nx^{n-1}$

Thus,

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Question

Find the average acceleration on the interval $0\leq t \leq 7$ if the velocity function is .

Answer

Average acceleration is defined as $a_{avg} = \frac{\Delta v}{\Delta t}$ .

Therefore, on the interval $0 \leq t \leq 7$ ,

$\a_{avg} = \frac{v(7)-v(0)}{7-0}\ a_{avg} = \frac{56 - (-35)}{7}\ a_{avg} = 13$

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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

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

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

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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