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Calculus 1 : How to find acceleration

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Example Questions

Example Question #341 : How To Find Acceleration

The position of a t=4 is given by the following functions:

p(t) = 2t + 3

Find the acceleration.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function and the derivative of the velocity function to get the acceleration function: p''(t) = v'(t) = a(t)

In this case, the position function is: p(t) = 2t + 3

Then take the derivative of the position function to get the velocity function: v(t) = 2

Then take the derivative of the velocity function to get the acceleration function: a(t) = 0

Then, plug t=4 into the acceleration function: a(4) = 0

Therefore, the answer is:

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Example Question #341 : Acceleration

Given the position function of an object in motion (in meters), find the acceleration of the object at t=2 seconds.

s(t)=-4.9t^2-8t+3

Possible Answers:

-4.9 m/s/s

-9.8 m/s/s

None of the other answers.

4.9 m/s/s

9.8 m/s/s

Correct answer:

-9.8 m/s/s

Explanation:

The acceleration of an object in motion at any given time is modeled by the second derivative of its position function. That is

s''(t)=a(t)=acceleration

s''(t)=d/dt(s'(t))=d/dt(-9.8t-8)=-9.8=a(t)

The acceleration of the object at t=2 seconds is then

a(2)=-9.8

In this case, the acceleration is constant across all values of t.

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Example Question #342 : How To Find Acceleration

Find the acceleration of a particle at t=5 given the following velocity function:

V(t)= 6t^2+2t+1

Possible Answers:

A(5)= 25

A(5)= 62

A(5)= 66

A(5)= 100

Correct answer:

A(5)= 62

Explanation:

The first step is to obtain the acceleration equation.

This is done by taking the first derivative of the velocity function,

V(t)= 6t^2+2t+1

using the power rule,

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

The acceleration equation

$A(t)=V'(t)= 2\cdot 6t^{2-1}+1\cdot 2t^{1-1}$

A(t)= 12t+2

can be used to find the acceleration at any given time.

In our case, plugging in 5 for t into the equation gives,

$\\A(5)= 12(5)+2 \\A(5)=60+2$

A(5)= 62 .

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Example Question #341 : How To Find Acceleration

Find the acceleration of a particle given its velocity function is v(t)= 2t^5+5t^2+300 .

Possible Answers:

a(t)= 10t^4 +10t

a(t)= 10t^5+10t

a(t)= 10t^4 +5t + 300

a(t)= 2t+5

Correct answer:

a(t)= 10t^4 +10t

Explanation:

The acceleration function a(t) can be found given any velocity function by taking its derivative.

Using the power rule

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

the derivative becomes,

$v'(t)= 5\cdot 2t^{5-1}+2\cdot 5t^{2-1}$

v'(t)=a(t)= 10t^4+10t .

Recall that the derivative of a constant is always zero.

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Example Question #342 : Acceleration

Find the acceleration function of a car who's velocity function is given by v(t)=2t^2+5 .

Possible Answers:

a(t)=2t+5

a(t)=4t^2

a(t)=4t+5

a(t)=4t

Correct answer:

a(t)=4t

Explanation:

The derivative of the velocity function is acceleration, and using the power rule

$\frac{\mathrm{d} }{\mathrm{d} x}(x^n)=nx^{n-1}$ ,

you get

$v'(t)=2\cdot 2t^{2-1}$

a(t)=4t .

Recall that the derivative of a constant is zero.

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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 #341 : How To Find Acceleration

Example Question #341 : Acceleration

Example Question #342 : How To Find Acceleration

Example Question #341 : How To Find Acceleration

Example Question #342 : Acceleration

All Calculus 1 Resources

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