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Calculus 1 : Acceleration

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Example Questions

Example Question #331 : How To Find Acceleration

A peach falls from a tree with a velocity given by the following function:

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

What is the acceleration of the peach at t=3?

Possible Answers:

$\frac{1}{2}$

$-\infty$

$-\frac{9}{4}$

$-\frac{3}{2}$

$\frac{3}{2}$

Correct answer:

$-\frac{3}{2}$

Explanation:

To find the acceleration of the peach, we must take the first derivative of the velocity function:

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

The derivative was found using the following rule:

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

Now, to find the velocity at the specific t, plug in this t value into the acceleration function:

$a(3)=-\frac{3}{2}$ .

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Example Question #331 : Acceleration

Find the acceleration of a jet engine at t=1 if its position is given by:

$p(t)=e^{\frac{1}{t}}$

Possible Answers:

e^2

Correct answer:

Explanation:

To find acceleration, we take two derivatives of position:

$p(t)=e^{\frac{1}{t}}$

By the chain rule:

$p'(t)=e^{\frac{1}{t}}*-\frac{1}{t^2}=-\frac{e^\frac{1}{t}}{t^2}$

To take the derivative of this, we need the quotient rule:

$p''(t)=-\frac{t^2*e^{\frac{1}{t}}*-\frac{1}{t^2}-2t*e^{\frac{1}{t}}}{t^4}=\frac{e^{\frac{1}{t}}(2t+1)}{t^4}$

At t=1 ,

$p''(t)=\frac{e^{\frac{1}{1}}(2*1+1)}{1^4}=3e$

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Example Question #721 : Spatial Calculus

If p(t) models the position of a guitar string as a function of time, find the acceleration function of the string.

p(t)=5sin(t)-10t

Possible Answers:

a(t)=5sin(t)-10

a(t)=-5sin(t)

a(t)=-5t

Correct answer:

a(t)=-5sin(t)

Explanation:

If p(t) models the position of a guitar string as a function of time, find the acceleration function of the string.

p(t)=5sin(t)-10t

We start with position and are asked to find acceleration. This means we are going to find the second derivative of p(t).

Recall that the derivative of sine is cosine, and the derivative of a linear term is a constant.

With that in mind, we get the following:

p'(t)=5cos(t)-10

p''(t)=-5sin(t)

So we get:

a(t)=-5sin(t)

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Example Question #333 : How To Find Acceleration

If p(t) models the position of a guitar string as a function of time, find the acceleration of the string when $t=\frac{5\pi}{6}$ .

p(t)=5sin(t)-10t

Possible Answers:

$0\frac{m}{s^2}$

$-5\frac{m}{s^2}$

$-2.5\frac{m}{s^2}$

$-7.5\frac{m}{s^2}$

Correct answer:

$-2.5\frac{m}{s^2}$

Explanation:

If p(t) models the position of a guitar string as a function of time, find the acceleration of the string when $t=\frac{5\pi}{6}$ .

p(t)=5sin(t)-10t

We start with position and are asked to find acceleration. This means we are going to find the second derivative of p(t).

Recall that the derivative of sine is cosine, and the derivative of a linear term is a constant.

With that in mind, we get the following:

p'(t)=5cos(t)-10

p''(t)=-5sin(t)

So we get:

a(t)=-5sin(t)

However, we need to find $a(\frac{5\pi}{6})$

$a(t)=-5sin(\frac{5\pi}{6})=-2.5$

So our answer is

$-2.5\frac{m}{s^2}$

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Example Question #332 : Acceleration

What is the acceleration function when x(t)=4t^2-t+3 ?

Possible Answers:

a(t)=8t

a(t)=0

$a(t)=\frac{1}{8}$

a(t)=8

a(t)=-8

Correct answer:

a(t)=8

Explanation:

To find acceleration, you must first find the velocity function, which is the derivative of the position function. To take the derivative, multiply the exponent by the coefficient and then subtract one from the exponent: v(t)=8t-1 . Then, take the derivative of that to find the acceleration: a(t)=8 .

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Example Question #334 : How To Find Acceleration

The position of a t=0 is given by the following functions:

$p(t) = ln(3t + \sin(3t))$

Find the acceleration.

Possible Answers:

Answer not listed

Correct answer:

Answer not listed

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: $p(t) = ln(3t + \sin(3t))$

Then take the derivative of the position function to get the velocity function: $v(t) = \frac{3+ 3\cos(3t)}{3t + \sin(3t)}$

Then take the derivative of the velocity function to get the acceleration function: $a(t) = -\frac{9 \sin(3t)}{3t + \sin(3t)} - \frac{(3 + 3\cos(3t))^2}{(3t + \sin(3t))^2}$

Then, plug t=0 into the acceleration function: $a(0) = -\frac{9 \sin(3({\color{Blue} 0}))}{3({\color{Blue} 0}) + \sin(3({\color{Blue} 0}))} - \frac{(3 + 3\cos(3({\color{Blue} 0})))^2}{(3({\color{Blue} 0}) + \sin(3({\color{Blue} 0})))^2}$

Therefore, the answer is: undefined

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Example Question #337 : Acceleration

The position of a t=2 is given by the following functions:

p(t) = 4t^2 - e^t

Find the acceleration.

Possible Answers:

Answer not listed

1.4

0.6

3.5

0.8

Correct answer:

0.6

Explanation:

In this case, the position function is: p(t) = 4t^2 - e^t

Then take the derivative of the position function to get the velocity function: v(t) = 8t - e^t

Then take the derivative of the velocity function to get the acceleration function: a(t) = 8 - e^t

Then, plug t=2 into the acceleration function: $a(2) = 8 - e^{\color{Blue} 2}$

Therefore, the answer is: 0.6

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Example Question #335 : How To Find Acceleration

The position of a t=0 is given by the following functions:

$p(t) = e^{3t-ln(7)}$

Find the acceleration.

Possible Answers:

$\frac{3}{}4$

$\frac{1}{8}$

$\frac{9}{7}$

Answer not listed

$\frac{6}{5}$

Correct answer:

$\frac{9}{7}$

Explanation:

In this case, the position function is: $p(t) = e^{3t-ln(7)}$ = $p(t) = \frac{e^{3t}}{7}$

Then take the derivative of the position function to get the velocity function: $v(t) = \frac{3e^{3t}}{7}$

Then take the derivative of the velocity function to get the acceleration function: $a(t) = \frac{9e^{3t}}{7}$

Then, plug t=0 into the acceleration function: $a(0) = \frac{9e^{3({\color{Blue} 0})}}{7}$

Therefore, the answer is: $\frac{9}{7}$

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Example Question #722 : Spatial Calculus

The position of a t=2 is given by the following functions:

$p(t) = - \cos(3t - 6)$

Find the acceleration.

Possible Answers:

Answer not listed

Correct answer:

Explanation:

In this case, the position function is: $p(t) = - \cos(3t - 6)$

Then take the derivative of the position function to get the velocity function: $v(t) = 3\sin(3t - 6)$

Then take the derivative of the velocity function to get the acceleration function: $a(t) = 9\cos(3t - 6)$

Then, plug t=2 into the acceleration function: $a(t) = 9\cos(3({\color{Blue} 2}) - 6)$

Therefore, the answer is:

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Example Question #333 : Acceleration

The position of a t=1 is given by the following functions:

p(t) = ln(4t^2 + 6t)

Find the acceleration.

Possible Answers:

6.2

1.2

Answer not listed

4.4

2.8

Correct answer:

2.8

Explanation:

In this case, the position function is: p(t) = ln(4t^2 + 6t)

Then take the derivative of the position function to get the velocity function: $v(t) = \frac{8t + 6}{4t^2+6t}$

Then take the derivative of the velocity function to get the acceleration function: $a(t) = \frac{8}{4t^2+6t} - \frac{(8t+6)^2}{(4t^2+6t)^2}$

Then, plug t=1 into the acceleration function: $a(1) = \frac{8}{4({\color{Blue} 1})^2+6({\color{Blue} 1})} - \frac{(8({\color{Blue} 1})+6)^2}{(4({\color{Blue} 1})^2+6({\color{Blue} 1}))^2}$

Therefore, the answer is: 2.8

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Calculus 1 : Acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #331 : How To Find Acceleration

Example Question #331 : Acceleration

Example Question #721 : Spatial Calculus

Example Question #333 : How To Find Acceleration

Example Question #332 : Acceleration

Example Question #334 : How To Find Acceleration

Example Question #337 : Acceleration

Example Question #335 : How To Find Acceleration

Example Question #722 : Spatial Calculus

Example Question #333 : Acceleration

All Calculus 1 Resources

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