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Calculus 1 : How to find acceleration

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Example Question #11 : Acceleration

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

Possible Answers:

Correct answer:

Explanation:

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

v(t)=40

The derivative of the velocity function is the acceleration function.

a(t)=0

The acceleration at any particular time is zero.

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Example Question #12 : Acceleration

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.

Possible Answers:

m/s^2

9.8 m/s^2

3.2 m/s^2

m/s^2

Correct answer:

m/s^2

Explanation:

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.

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

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

a(t)=v'(t)=s''(t)=8

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Example Question #13 : Acceleration

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.

Possible Answers:

a(t)=2t+16

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

$a(t)=\frac{1}{2}t-16$

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

$a(t)=16t+\frac{1}{3}$

Correct answer:

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

Explanation:

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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Example Question #14 : Acceleration

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 a(0)=0.1 , find the acceleration of the object at t=3 seconds.

Possible Answers:

0.7 m/s^2

0.3 m/s^2

m/s^2

0.1 m/s^2

Correct answer:

0.7 m/s^2

Explanation:

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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Example Question #6 : Derivatives

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 t=1 second.

Possible Answers:

m/s^2

$\frac{5}{4}$ m/s^2

m/s^2

$\frac{15}{2}$ m/s^2

$\frac{3}{2}$ m/s^2

Correct answer:

$\frac{3}{2}$ m/s^2

Explanation:

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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Example Question #15 : Acceleration

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 t=2 seconds.

Possible Answers:

$\frac{3}{7}$ m/s^2

m/s^2

$\frac{1}{14}$ m/s^2

$\frac{6}{7}$ m/s^2

$\frac{23}{14}$ m/s^2

Correct answer:

$\frac{23}{14}$ m/s^2

Explanation:

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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Example Question #16 : Acceleration

The position vector of an object moving in a plane is given by r(t) = t^4i+3t^2j . Find the object's acceleration when t=3 .

Possible Answers:

114ij

114

108i+6j

102ij

Correct answer:

108i+6j

Explanation:

To find the acceleration, we must differentiate the position vector twice. Differentiating the position vector once gives the velocity vector:

$v(t) = (1\cdot 4)t^{4-1}i+(3\cdot 2)t^{2-1}j = 4t^3i+6tj$

Differentiating the second time gives acceleration:

$a(t) = (4\cdot 3)t^{3-1}i+(6\cdot 1)t^{1-1}j = 12t^2i+6j$

Using 3 as the value for gives

$a(3) = 12(3)^2i+6j = 12\cdot 9i + 6j = 108i+6j$

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Example Question #17 : Acceleration

If the position of a particle is: s(t) = 3t^3 + 10t , What is the acceleration at t = 3 ?

Possible Answers:

Correct answer:

Explanation:

To find the acceleration equation of the particle, we can differentiate the position equation twice.

Differentiating once gives the velocity equation:

v(t) = 9t^2+10

Differentiating again give the acceleration equation:

a(t) = 18t

Using 3 as the value for ,

a(3) = 18(3) = 54

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Example Question #18 : Acceleration

What is the acceleration at time t=4secs when the acceleration is given by a(t)=4t . is time in seconds.

Possible Answers:

a(4)=12

a(4)=8

a(4)=18

a(4)=4

a(4)=16

Correct answer:

a(4)=16

Explanation:

To find the acceleration at time 4 seconds, we simply subsitute in 4 seconds into the acceleration function.

$a(4)=4\cdot 4$

a(4)=16

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Example Question #19 : Acceleration

The velocity function is v(t)=4t .

What is the acceleration function?

Possible Answers:

a(t)=4t^2

a(t)=4t

a(t)=4

a(t)=2t^2

a(t)=2

Correct answer:

a(t)=4

Explanation:

To find the acceleration function we take the derivative of the velocity function

So v(t)=4t will turn into:

a(t)=4

because of Power Rule,

$p(t)=a^n \rightarrow v(t)=na^{n-1}$

$v(t)=4\cdot 1 t^{1-1}=4t^0=4(1)=4$ .

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Calculus 1 : How to find acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #11 : Acceleration

Example Question #12 : Acceleration

Example Question #13 : Acceleration

Example Question #14 : Acceleration

Example Question #6 : Derivatives

Example Question #15 : Acceleration

Example Question #16 : Acceleration

Example Question #17 : Acceleration

Example Question #18 : Acceleration

Example Question #19 : Acceleration

All Calculus 1 Resources

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