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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #22 : Acceleration

Given the vector position:

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

Find the expression of the velocity.

Possible Answers:

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

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

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

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

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

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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Example Question #31 : How To Find Acceleration

Suppose the position function of a particle can be traced by s(t)=2t^4-3t^3 , where is in seconds. What is the acceleration of the particle at t=3 ?

Possible Answers:

162

270

135

Correct answer:

162

Explanation:

To find the acceleration of the particle given the position function, take the 2nd derivative of the position function using the Power Rule.

s(t)=2t^4-3t^3

s'(t)=v(t)=8t^3-9t^2

a(t)=s''(t)=v'(t)=24t^2-18t

Substitute t=3 into the acceleration function to find the acceleration at the given time.

a(3)=24(3)^2-18(3)=216-54=162

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Example Question #31 : How To Find Acceleration

A car is driving north on a highway at a constant velocity of mph. What is the acceleration after an hour?

Possible Answers:

Correct answer:

Explanation:

If a car is travelling north at constant velocity 60 mph, it's possible to write a velocity function for this vehicle, where is time in hours.

v(t)=60

To find the acceleration, take the derivative of the velocity function.

a(t)=v'(t)=0

The acceleration after an hour, or any time , is zero.

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Example Question #31 : How To Find Acceleration

The velocity of an object is given by the equation $f(t) = \sin^3(2t)$ . What is the equation of acceleration of the object?

Possible Answers:

$12\sin^2(2t)\sin(2t)\cos(2t)$

$3\sin^2(2t)\cos(2t)$

$6\sin^2(2t)\cos(2t)$

$6\sin^2(2t)$

Correct answer:

$6\sin^2(2t)\cos(2t)$

Explanation:

The acceleration of the object can be found by taking the derivative of the velocity equation.

$f(t) = \sin^3(2t)$

The acceleration equation is then

$f'(t) = 3\sin^2(2t)\cos(2t)(2) = 6\sin^2(2t)\cos(2t)$

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Example Question #31 : How To Find Acceleration

If a car's velocity goes from mph to mph in seconds, what is the average acceleration?

Possible Answers:

$\frac{1}{3}$

Correct answer:

Explanation:

Write the formula for average acceleration.

$a=\frac{\Delta v}{t}=\frac{v_{f}-v_{i}}{t}$

The initial velocity is 50 mph, and the final velocity is 20 mph.

Substitute all givens and solve for the acceleration.

$a=\frac{20-50}{10}=\frac{-30}{10}=-3$

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Example Question #31 : How To Find Acceleration

The position of an object is describe by the function s(t)= 9cos(2t) . What is the acceleration at $t=\pi$ ?

Possible Answers:

2.25

-2.25

Acceleration cannot be negative.

Correct answer:

Explanation:

To find the acceleration function, derive s(t)= 9cos(2t) twice. To do this, use the Power Rule along with the Chain Rule.

v(t)=-18sin(2t)

a(t)=-36cos(2t)

Substitute $t=\pi$ into the acceleration function and solve.

$a(t)=-36cos(2\cdot \pi)= -36(1)= -36$

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Example Question #32 : How To Find Acceleration

Consider the position function {}p(t) , which describes the positon of an oxygen molecule.

p(t)=14t^3-14t^2+14t-14

Find the function which models the acceleration of the particle.

Possible Answers:

a(t)=84-28t

a(t)=84

a(t)=84t-28

a(t)=84t

a(t)=-28

Correct answer:

a(t)=84t-28

Explanation:

Recall that velocity is the first derivative of position and acceleration is the second derivative of position.

So given:

p(t)=14t^3-14t^2+14t-14

Apply the power rule to each term to find the velocity.

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

v(t)=p'(t)=42t^2-28t+14

Applying the power rule a second time we arrive at the acceleration function.

a(t)=p''(t)=84t-28

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Example Question #32 : How To Find Acceleration

Consider the position function {}p(t) , which describes the positon of an oxygen molecule.

p(t)=14t^3-14t^2+14t-14

Find the acceleration of the particle after seconds.

Possible Answers:

224

Correct answer:

224

Explanation:

Recall that velocity is the first derivative of position and acceleration is the second derivative of position.

So given:

p(t)=14t^3-14t^2+14t-14

v(t)=p'(t)=42t^2-28t+14

a(t)=p''(t)=84t-28

Find a(3) to find the acceleration after 3 seconds:

$a(3)=84\cdot 3-28=224$

So our acceleration after three seconds is 224 units per second

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Example Question #421 : Calculus

Find the acceleration of an object at t=5 if the velocity function is v(t)=ln(t) .

Possible Answers:

$-\frac{1}{5}$

$\frac{1}{ln(5)}$

$\frac{1}{5}$

ln(5)

Correct answer:

$\frac{1}{5}$

Explanation:

To find the acceleration of an object from the velocity function, simply take the derivative of the velocity function and substitute t=5 into the acceleration function.

Take the derivative of v(t) .

$a(t)= \frac{d(v(t))}{dt}=\frac{1}{t}$

Substitute the time to find acceleration.

$a(t)=\frac{1}{5}$

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Example Question #31 : How To Find Acceleration

The displacement of an object at time is defined by the function f(t) = 2t +1 . What is the acceleration at t=3 ?

Possible Answers:

Correct answer:

Explanation:

The acceleration is the 2nd derivative of the displacement. Knowing the equation of displacement, f(t) = 2t +1 we can use the power rule to differentiate.

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

Differentiating the equation gives us the velocity,

f'(t) = 2

Differentiating the equation a second time gives us the acceleration.

f''(t) = 0

Therefore, the answer is 0.

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #22 : Acceleration

Example Question #31 : How To Find Acceleration

Example Question #31 : How To Find Acceleration

Example Question #31 : How To Find Acceleration

Example Question #31 : How To Find Acceleration

Example Question #31 : How To Find Acceleration

Example Question #32 : How To Find Acceleration

Example Question #32 : How To Find Acceleration

Example Question #421 : Calculus

Example Question #31 : How To Find Acceleration

All Calculus 1 Resources

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