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Calculus 1 : Acceleration

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Example Question #21 : Acceleration

What is the acceleration function when the velocity function is

v(t)=0 ?

Possible Answers:

a(t)=1

a(t)=2

a(t)=0

a(t)=t^2

a(t)=t

Correct answer:

a(t)=0

Explanation:

Since the velocity function is . The acceleration function is also , since the derivative of is .

Therefore,

a(t)=0 .

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Example Question #7 : Derivatives

If h(t) models the distance of a projectile as a function of time, find the acceleration of the projectile at t=6 .

$h(t)=\frac{4x^3}{3}+\frac{4.5x^2}{3}+3x-7$

Possible Answers:

$51\frac{m}{s^2}$

$165\frac{m}{s^2}$

$48\frac{m}{s^2}$

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

Correct answer:

$51\frac{m}{s^2}$

Explanation:

We are given a function dealing with distance and asked to find an acceleration. recall that velocity is the first derivative of position and acceleration is the derivative of velocity. Find the second derivative of h(t) and evaluate at t=6.

$h(t)=\frac{4t^3}{3}+\frac{4.5t^2}{3}+3t-7$

h'(t)=4t^2+3t+3

h''(t)=8t+3

h''(6)=8*6+3=48+3=51

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Example Question #22 : Acceleration

Function v(t) 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 equation that models that particle's acceleration over time.

Possible Answers:

$\small \small \small \small \small \small \small \small a(t)=5t^4+4t^2-\frac{14t}{3}$

$\small \small \small \small \small \small \small a(t)=5t^4+12t^2-\frac{14}{3}$

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

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

$\small \small \small \small \small \small \small a(t)=20t^3+36t^2-\frac{14}{3}$

Correct answer:

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

Explanation:

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

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

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Example Question #23 : Acceleration

Function v(t) 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 t=5 seconds.

Possible Answers:

$\small \small 4625$

$\small 3125\frac{2}{3}$

$\small 3125$

$\small 23\frac{1}{3}$

$\small 4601\frac{2}{3}$

Correct answer:

$\small 4601\frac{2}{3}$

Explanation:

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

$\small \small \small \small \small \small \small a(t)=5*5^4+12*5^3-\frac{14*5}{3}=3125+1500-\frac{70}{3}=4601\frac{2}{3}$

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Example Question #24 : Acceleration

Consider the velocity function modeled in meters per second by v(t) .

v(t)=4t^3-5t^2+6

Find the accelaration after seconds of a particle whose velocity is modeled by v(t) .

Possible Answers:

$25\frac{m}{s^2}$

$255\frac{m}{s^2}$

$350\frac{m}{s^2}$

$250\frac{m}{s^2}$

$10\frac{m}{s^2}$

Correct answer:

$250\frac{m}{s^2}$

Explanation:

Recall that acceleration is the first derivative of velocity. So, to find acceleration we need to find and evaluate the following:

v'(5)

So, if

v(t)=4t^3-5t^2+6

Then,

$v'(t)=3\cdot4t^{3-1}-2\cdot5t^{2-1}=12t^2-10t$

$v'(5)=12\cdot 5^2-10\cdot 5=250$

So our acceleration is $250\frac{m}{s^2}$ .

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Example Question #25 : Acceleration

Consider the following position function:

p(t)=6t^4-5t^3+6t-18

Find the acceleration after seconds of a particle whose position is given by {}p(t) .

Possible Answers:

$3318 \frac{m}{s^2}$

$33 \frac{m}{s^2}$

$3339 \frac{m}{s^2}$

$318 \frac{m}{s^2}$

$33.18 \frac{m}{s^2}$

Correct answer:

$3318 \frac{m}{s^2}$

Explanation:

Recall that acceleration is the second derivative of position, so we need p''(7).

p(t)=6t^4-5t^3+6t-18

Taking the first derivative we get:

p''(t)=72t^2-30t

Taking the second derivative and plugging in 7 we get:

$p''(7)=72\cdot7^2-30\cdot7=3318$

So our acceleration after 7 seconds is

$3318 \frac{m}{s^2}$ .

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Example Question #26 : Acceleration

Given that the position of a particle is known to be:

$\vec{OM}=(tsin(t),tcos(t),1)$

Find the acceleration of the particle.

Possible Answers:

(-2cos(t)-tsin(t),-tcos(t)-2sin(t),0)

(2cos(t)-tsin(t),-tcos(t)-2sin(t),0)

(2cos(t)-tsin(t),-tcos(t)+2sin(t),0)

(2cos(t)-tsin(t),tcos(t)-2sin(t),0)

(2cos(t)+tsin(t),-tcos(t)-2sin(t),0)

Correct answer:

(2cos(t)-tsin(t),-tcos(t)-2sin(t),0)

Explanation:

To find the expression of the acceleration we need to differentiate the expression of the position vector twice with respect to time.

Note that we do differentiation componentwise.

We have the following:

(tsin(t))'=sin(t)+tcos(t)

(tcos(t))'=cos(t)-tsin(t) .

We differentiate one more time:

(tsin(t))''=2cos(t)-tsin(t)

(tcos(t))''=-tcos(t)-2sin(t)

and the third component is simply zero since it is a constant.

This gives us the expression for the acceleration.

$\vec{a}=(2cos(t)-tsin(t),-tcos(t)-2sin(t),0)$

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Example Question #27 : Acceleration

A planet is moving along the path:

$\vec{r}=(artan(t),ln(t),e^{t})$

What is its acceleration?

Possible Answers:

$\left(\frac{2t}{(1+t^2)^2},-\frac{1}{t^2},e^{t}\right)$

$\left(-\frac{2t}{(1+t^2)^2},-\frac{1}{t^2},e^{t}\right)$

$\left(-\frac{2t}{(1+t^2)^2},-\frac{1}{t^3},e^{t}\right)$

$\left(-\frac{2t}{(1+t^2)^2},\frac{1}{t^2},e^{t}\right)$

$\left(-\frac{2t}{(1+t^3)^2},-\frac{1}{t^2},e^{t}\right)$

Correct answer:

$\left(-\frac{2t}{(1+t^2)^2},-\frac{1}{t^2},e^{t}\right)$

Explanation:

To find the correct expression of the acceleration, we will need to differentiate the expression of the position twice with respect to time.

We recall the following:

$arctan(t)'=\frac{1}{1+t^2}\rightarrow (artan(t))''=\left(-\frac{2t}{(1+t^2)^2}\right)$

$ln(t)'=\frac{1}{t}\rightarrow (ln(t))''=-\frac{1}{t^2}$

and

$e(t)'=e^t \rightarrow (e^{t})''=e^{t}$ .

This gives us

$\vec{a}=\left(-\frac{2t}{(1+t^2)^2},-\frac{1}{t^2},e^{t}\right)$

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Example Question #28 : Acceleration

Assume that a particle is moving along the path given by the equations :

$\left\{\begin{matrix}x(t)=t+1\\y(t)=ln(t)+1 \\z(t)=tln(t)+1 \end{matrix}\right.$

What is the acceleration of the particle?

Possible Answers:

$\left(0,\frac{-1}{t^3},\frac{1}{t}\right)$

$\left(0,\frac{1}{t^2},\frac{1}{t}\right)$

$\left(0,\frac{-1}{t^2},\frac{1}{t}\right)$

$\left(t,\frac{-1}{t^2},\frac{1}{t}\right)$

$\left(0,\frac{1}{t^2},\frac{1}{t^2}\right)$

Correct answer:

$\left(0,\frac{-1}{t^2},\frac{1}{t}\right)$

Explanation:

To find the expression of the acceleration we need to find the second derivative of each of the components with respect to time.

Note that we have the following:

The first component is zero since its second derivative is zero.

$x(t)=(t+1)'=1 \rightarrow 1'=0$

$y(t)=(ln(t)+1)'=\frac{1}{t}\rightarrow (ln(t)+1)''=-\frac{1}{t^2}$

$z(t)=(t ln(t))'=\frac{t}{t}+ln(t)\rightarrow (t ln(t))''=\frac{1}{t}$

This gives the required expression.

$\left(0,\frac{-1}{t^2},\frac{1}{t}\right)$

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Example Question #29 : Acceleration

Given the vector position:

$\vec{r}=(tcos(t),t, tsin(t)).$

Find the expression of the velocity.

Possible Answers:

$\vec{v}=(cos(t)-tsin(t),1,cos(t)+tcos(t))$

$\vec{v}=(cos(t)-tsin(t),1,sin(t)+tcos(t))$

$\vec{v}=(cos(t)-tcos(t),1,sin(t)+tcos(t))$

$\vec{v}=(cos(t)+tsin(t),1,sin(t)+tcos(t))$

$\vec{v}=(cos(t)-tsin(t),1,sin(t)-tcos(t))$

Correct answer:

$\vec{v}=(cos(t)-tsin(t),1,sin(t)+tcos(t))$

Explanation:

All we need to do to find the components of the velocity is to differentiate the components of the position vector with respect to time.

We have :

(tcos(t))'=cos(t)-tsin(t)

t'=1

(tsin(t))'=sin(t)+tcos(t)

Collecting the components we have :

$\vec{v}=(cos(t)-tsin(t),1,sin(t)+tcos(t))$

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Calculus 1 : Acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #21 : Acceleration

Example Question #7 : Derivatives

Example Question #22 : Acceleration

Example Question #23 : Acceleration

Example Question #24 : Acceleration

Example Question #25 : Acceleration

Example Question #26 : Acceleration

Example Question #27 : Acceleration

Example Question #28 : Acceleration

Example Question #29 : Acceleration

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