Spatial Calculus - CLEP Calculus

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Question

What is the instantaneous acceleration at time t = π/2 of a particle whose positional equation is represented by s(t) = 5sin(t/2 + π)?

Answer

The instantaneous acceleration is represented by the second derivative of the positional equation. Let's first calculate the velocity then the acceleration:

s(t) = 5sin(t/2 + π)?

v(t) = s'(t) = (5/2)cos(t/2 + π)

a(t) = v'(t) = s''(t) = –(5/4)sin(t/2 + π)

We know that sin(x) = –sin(x + π); therefore, sin(t/2 + π) = –sin(t/2). Let's rewrite a(t) as (5/4)sin(t/2).

The acceleration is therefore a(π/2) = (5/4)sin(π/2/2) = (5/4)sin(π/4) = 5/4 * (1/√2) = 4/(4√2) = 5√2/8

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Question

The velocity of a particle is given by the function $v(t)=1.6^{t^2}$ . Find the acceleration of the particle at time

Answer

Acceleration of a particle can be found by taking the derivative of the velocity function with respect to time. Recall that a derivative gives the rate of change of some parameter, relative to the change of some other variable. When we take the derivative of velocity with respect to time, we are evaluating how velocity changes over time; i.e acceleration! This is just like finding velocity by taking the time derivative of the position function.

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

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

To take the derivative of the function

$v(t)=1.6^{t^2}$

We'll need to make use of the following derivative rule:

Derivative of an exponential:

Note that u may represent large functions, and not just individual variables!

Using the above properties, the acceleration function is

$a(t)=2t1.6^{t^2}ln(1.6)$

$a(2)=2(2)1.6^{2^2}ln(1.6)=12.3$

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Question

The relative displacement (in meters) of a particle at time t is defined by the function f(t) = 4t2 * ln(t)

What is the instantaneous acceleration of the particle at t = 3.5?

Answer

The instantaneous acceleration is found by taking the 2nd derivative of the function and applying thereto the desired variable parameter. First let us calculate the 1st derivative:

f(t) = 4t2 * ln(t) will require us to apply the product rule; therefore:

f'(t) = 8t * ln(t) + 4t2 * (1/t), which reduces to: 8tln(t) + 4t

Taking the second derivative (applying the product rule to the first value), we get:

f''(t) = 8 ln(t) + 8t * (1/t) + 4, which simplifies to: 8 * ln(t) + 8 + 4 = 8 * ln(t) + 12

The instantaneous acceleration at t = 3.5 is found by solving:

f''(3.5) = 8 * ln(3.5) + 12 = 22.02210374796294 (approx.) or 22.02

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Question

What is the instantaneous acceleration at time t = 4 of a particle whose positional equation is represented by s(t) = 5t3 – 24t2 + 44?

Answer

The instantaneous acceleration is represented by the second derivative of the positional equation. Let's first calculate the velocity then the acceleration:

v(t) = s'(t) = 15t2 – 48t

a(t) = v'(t) = s''(t) = 30t – 48

a(4) = 30 * 4 – 48 = 72

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Question

What is the instantaneous acceleration at time t = 25 of a particle whose positional equation is represented by s(t) = –44t2 + 70√t?

Answer

The instantaneous acceleration is represented by the second derivative of the positional equation. Let's first calculate the velocity then the acceleration. Begin by rewriting s(t) to make the differentiation easier:

s(t) = –44t2 + 70√t = –44t2 + 70(t)1/2

v(t) = s'(t) = –88t + 70 * (1/2) * t–1/2 = –88t + 35t–1/2

a(t) = v'(t) = s''(t) = –88 - 70t–3/2 = –88 -70/(t√t)

a(5) = –88 -70/(25√25) = -88 - 70/(5 * 5) = –88 - 70/25 = –88.82

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Question

The position vector of an object moving in a plane is given by r(t)=3t4i + 6t2j. Find its acceleration when t=1.

Answer

To find acceleration at time t, we have to differentiate the position vector twice.

Differentiating the first time gives the velocity:

v(t) = r'(t) = 12t3i + 12tj

Differentiating a second time gives the accelaration:

a(t) = r''(t) = 36t2i + 12j

Plug in t=1 to solve for the final answer:

a(1) = r''(1) = 36i + 12j

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Question

What is the instantaneous acceleration at time t = π of a particle whose positional equation is represented by s(t) = 5sin(4t) + cos(2t)?

Answer

The instantaneous acceleration is represented by the second derivative of the positional equation. Let's first calculate the velocity then the acceleration:

s(t) = 5sin(4t)+ cos(2t)?

v(t) = s'(t) = 20cos(4t) – 2sin(2t)

a(t) = v'(t) = s''(t) = –80sin(4t) – 4cos(2t)

a(π) = –80sin(4π) – 4cos(2π) = –80sin(0) - 4cos(0) = 0 – 4 = –4

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Question

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

What is the acceleration at time ?

Answer

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

$a = \frac{dv}{dt}=\frac{d}{dt}(-3sin(t))=-3cos(t)$

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Question

Find the acceleration of a particle with a position given by at time

Answer

Acceleration is given by the second derivative of position. So for this equation, the second derivative is . The acceleration is constant through time, so the answer is simply 8.

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Question

Find the acceleration of a particle at any given time if its velocity is given by ?

Answer

The acceleration is the derivative of the velocity function. For the sinusoid component, the chain rule must be used to get the derivative.

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Question

Jerk is defined as the rate of change of acceleration. Find the function of jerk with respect to time for a particle with position given by

Answer

Since jerk is the rate of change of acceleration, and acceleration is the second derivative of position, the answer is to find the third derivative of position. Use the chain rule each time.

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Question

The position of a particle is given by $s(t)=e^{t^2}+t^2$ . Find the acceleration of the particle when .

Answer

The acceleration of a particle is given by the second derivative of the position function. We are given the position function as

$s(t)=e^{t^2}+t^2$ .

The first derivative (the velocity) is given as

$\frac{d(e^{t^2}+t^2)}{dt}=2te^{x^2}+2t$ .

The second derivative (the acceleration) is the derivative of the velocity function. This is given as

$\frac{d(2te^{t^2}+2t)}{dt}=2e^{t^2}+4t^2e^{t^2}+2$ .

Evaluating this at gives us the answer. Doing this we get

$a(1)=2e^{1^2}+41^2e^{1^2}+2=6e+2=2(3e+1)$ .

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Question

Suppose a particle travels in a circular motion in the xy-plane with

$x(t) = R \cos(\omega t)$

$y(t) = R \sin(\omega t)$

for some constants $\omega, R$ . Notice that this is circular because .

What is the total magnitude of the acceleration, $a = \sqrt{(x'')^2 + (y'')^2}$ ?

Answer

We know that and so the acceleration components are:

$x''(t) = - R \omega^2 \cos(\omega t)$

$y''(t) = - R \omega^2 \sin(\omega t)$

Plugging these into the formula for the acceleration and again recognizing that $\sin^2(\theta) + \cos^2(\theta) = 1$ , we get $a^2 = R^2 \omega^4$ , or, after the square root, $a = R \omega^2$ .

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Question

A car is moving at a constant speed of miles per hour. What is the acceleration after hours?

Answer

The car is moving at a constant speed of 40 miles per hour, so the velocity function is:

The derivative of the velocity function is the acceleration function.

The acceleration at any particular time is zero.

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Question

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

$v(t)=\frac{3}{2}t^2-\frac{1}{3}t+16$

Find the equation for the acceleration of the object.

Answer

Acceleration is the derivative of velocity, so in order to find the equation for the object's acceleration, we must take the derivative of the equation for its velocity:

$v(t)=\frac{3}{2}t^2-\frac{1}{3}t+16$

We will use the power rule to find the derivative which states:

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

$a(t)=v'(t)=\frac{3}{2}\cdot 2t^1-\frac{1}{3}t^0=3t-\frac{1}{3}$

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Question

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

$s(t)=4t^2-\frac{11}{3}t-7$

Find the acceleration of the object.

Answer

Velocity is the derivative of position, and acceleration is the derivative of velocity, so acceleration is the second derivative of position. With that in mind, all we have to do to find the acceleration of the object is take the derivative of the equation for its position twice.

$s(t)=4t^2-\frac{11}{3}t-7$

$v(t)=s'(t)=8t-\frac{11}{3}$

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Question

Function gives the velocity of a particle as a function of time.

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Find the acceleration (in meters per second per second) of the particle at seconds.

Answer

Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.

To derive a polynomial, simply decrease each exponent by one and bring the original number down in front to multiply.

So this

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Becomes:

$\small \small \small \small \small v'(t)=5t^4+12t^3-\frac{14t}{3}$

So our acceleration is given by

$\small \small \small \small \small \small a(t)=5t^4+12t^3-\frac{14t}{3}$

Now, to find the acceleration at 5 seconds, we need to plug in 5 for t

$\small \small \small \small \small \small \small a(t)=55^4+125^3-\frac{14*5}{3}=3125+1500-\frac{70}{3}=4601\frac{2}{3}$

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Question

The acceleration of an object is given by the folowing indefinite integral:

$a(t)=\int \left(\frac{1}{5}t-\frac{1}{10}\right)dt$

If , find the acceleration of the object at seconds.

Answer

In order to find a(3), our first step is to evaluate the integral in the equation for acceleration:

$a(t)=\int \left(\frac{1}{5}t-\frac{1}{10}\right)dt=\frac{1}{10}t^2-\frac{1}{10}t+C$

Now we use the initial acceleration, a(0)=0.1, to solve for the constant C:

$a(t)=\frac{1}{10}t^2-\frac{1}{10}t+C\rightarrow a(0)=0+0+C=0.1\rightarrow C=0.1$

So if C=0.1, then our final equation for acceleration is as follows, which we can then plug t=3 into to find the acceleration of the object after 3 seconds, a(3):

$a(t)=\frac{1}{10}t^2-\frac{1}{10}t+0.1$

$a(3)=\frac{1}{10}(3)^2-\frac{1}{10}(3)+0.1=0.9-0.3+0.1=0.7$

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Question

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

$s(t)=\frac{5}{4}t^3-3t^2+\frac{1}{2}t-6$

Find the acceleration of the object at second.

Answer

Acceleration is the second derivative of position, so we must first find the second derivative of the equation for position:

$s(t)=\frac{5}{4}t^3-3t^2+\frac{1}{2}t-6$

$v(t)=s'(t)=\frac{15}{4}t^2-6t+\frac{1}{2}$

$a(t)=v'(t)=s''(t)=\frac{15}{2}t-6$

Now we can plug in t=1 to find the acceleration of the object after 1 second:

$a(1)=\frac{15}{2}(1)-6=\frac{3}{2}$

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Question

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

$v(t)=\frac{3}{7}t^2-\frac{1}{14}t+5$

Find the acceleration of the object at seconds.

Answer

Acceleration is the derivative of velocity, so we must take the derivative of the given equation to find an equation for acceleration:

$v(t)=\frac{3}{7}t^2-\frac{1}{14}t+5$

$a(t)=v'(t)=\frac{6}{7}t-\frac{1}{14}$

Now we can plug in t=2 to find the acceleration of the object at 2 seconds:

$a(2)=\frac{6}{7}(2)-\frac{1}{14}=\frac{12}{7}-\frac{1}{14}=\frac{24}{14}-\frac{1}{14}=\frac{23}{14}$

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