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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 #101 : How To Find Distance

Find the position at t=2 given the following velocity function.

v(t)=2t+3

Possible Answers:

Correct answer:

Explanation:

To solve, simply integrate v(t) to find p(t) , the position function, and then plug in t=2 .

$\int_{0}^{2}v(t) \ dt=\int_{0}^{2}2t+3 \ dt=t^2+3t \Big |_{0}^{2}=(2^2+3(2))-(0^2+3(0))=4+6=10$

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Example Question #101 : How To Find Distance

The velocity of a particle is given by the function $v(t)=12t^2e^{t^3}$ . How far does the particle travel over the interval of time t=[1,2]

Possible Answers:

4812.91

410.32

28701.66

189351.51

11912.96

Correct answer:

11912.96

Explanation:

Velocity is the time derivative of position, and by that token position can be found by integrating a known velocity function with respect to time:

$p(t)=\int v(t)dt$

Now, if this integral were to be taken over an interval of time t=[a,b] , this will give a finite value, a change in position, i.e. a distance travelled:

$d=\int_a^b v(t)dt$

For the velocity function

$v(t)=12t^2e^{t^3}$

The distance travelled can be found via knowledge of the following derivative properties:

Derivative of an exponential:

d[e^u]=e^udu

The distance travelled over t=[1,2] is:

$d=4e^{t^3}|_1^2$

$d=4e^{2^3}-4e^{1^3}$

d=11912.96

Pretty speedy.

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Example Question #102 : How To Find Distance

The velocity of a particle is given by the function $v(t)=\frac{6t}{\cos^2(t^2)}$ . How far does the particle travel over the interval of time t=[0,0.4] ?

Possible Answers:

0.139

0.338

2.115

0.512

0.484

Correct answer:

0.484

Explanation:

Velocity is the time derivative of position, and by that token position can be found by integrating a known velocity function with respect to time:

$p(t)=\int v(t)dt$

Now, if this integral were to be taken over an interval of time t=[a,b] , this will give a finite value, a change in position, i.e. a distance travelled:

$d=\int_a^b v(t)dt$

For the velocity function

$v(t)=\frac{6t}{\cos^2(t^2)}$

The distance travelled can be found via knowledge of the following derivative properties:

Trigonometric derivative:

$d[\tan(u)]=\frac{du}{\cos^2(u)}$

The distance travelled over t=[0,0.4] is:

$d=3\tan(t^2)\Big|_0^{0.4}$

$d=3\tan(0.4^2)-3\tan(0^2)$

d=0.484

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Example Question #103 : How To Find Distance

The velocity of a particle is given by the function $v(t)=3t^2\cos(t^3)-8t\sin(t^2)$ . How far does the particle travel over the interval of time t=[2,5] ?

Possible Answers:

4.97

4.45

-0.98

-2.00

1.19

Correct answer:

4.97

Explanation:

Velocity is the time derivative of position, and by that token position can be found by integrating a known velocity function with respect to time:

$p(t)=\int v(t)dt$

Now, if this integral were to be taken over an interval of time t=[a,b] , this will give a finite value, a change in position, i.e. a distance travelled:

$d=\int_a^b v(t)dt$

For the velocity function

$v(t)=3t^2\cos(t^3)-8t\sin(t^2)$

The distance travelled can be found via knowledge of the following derivative properties:

Trigonometric derivative:

$d[\sin(u)]=\cos(u)du$

$d[\cos(u)]=-\sin(u)du$

The distance travelled over t=[2,5] is:

$d=\sin(t^3)+4\cos(t^2)\Big|_2^5$

$d=(\sin(5^3)+4\cos(5^2))-(\sin(2^3)+4\cos(2^2))$

d=4.97

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Example Question #102 : How To Find Distance

A young businessman goes to New York to try and find a job. When he gets to New York, he goes to the tallest building he can find. When he's at the top, he drops a penny off the side. 15 seconds later he hears someone yell "OW!!" and looks down to see a man on his balcony rubbing his head in pain.

Assuming that the acceleration due to gravity is equal to 10 m/s², and that the initial velocity was equal to zero, how far did the penny travel before hitting the poor guy in the head?

Possible Answers:

$850\ m$

$1000\ m$

$1250\ m$

$1125\ m$

$750\ m$

Correct answer:

$1125\ m$

Explanation:

The acceleration of the penny is equal to 10 m/s², meaning that every second that passes by, the velocity will increase by 10 m/s. In order to find a formula for the velocity of the penny, you must find the anti-derivative of the acceleration. Taking the anti-derivative of the veloctity will then give you a formula for the position of the penny.

a(t) = f''(t) = 10 m/s^2

v(t) = f'(t) = 10t + C .

Since the initial velocity was zero, you can use this initial condition to solve for the constant C. Setting 10t + C equal to zero, C is solved to be equal to 0.

Therefore, v(t) = 10t .

This makes sense, since every second that passes by, the velocity will increase by 10.

p(t) = f(t) = 5t^2+C .

The final integration (anti-derivative) yields this formula. Since we are only trying to find the distance traveled, and not its exact postion, the constant C is equal to zero.

Therefore, p(t) = 5t^2 m .

In order to find the total distance traveled, plug in the amount of time that passed into the postion formula you just found!

p(15) = 5(15^2) = 1125 meters traveled.

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Example Question #841 : Calculus

A gohper tortoise is walking down a straight road. The tortoise's velocity, measured in m/s, is modelled by the equation:

v=.02t+.05

How many meters has the gopher tortoise travelled after 20 seconds?

Possible Answers:

$8\ meters$

$6\ meters$

$5\ meters$

$4\ meters$

Correct answer:

$5\ meters$

Explanation:

To find distance traveled, we will integrate the velocity function:

$\int(.02t+.05)dt$

To integrate use the following rule,

$\int x^n=\frac{x^{n+1}}{n+1}$

Thus we get the following.

$=\frac{.02}{2}t^2+.05t$

=.01(t^2+5t)

Then, plugging in t=20

.01(20^2+5(20))=.01(500)=5

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Example Question #111 : Distance

Find the distance function given the following velocity function:

v(x)=2x-1

Possible Answers:

d(x)=0

d(x)=2

d(x)=x^2-x

d(x)=x^2-x+C

Correct answer:

d(x)=x^2-x+C

Explanation:

To solve, you must integrate v(t) to find d(t) once using the power rule for integrals.

$\int x^ndx=\frac{1}{n+1}x^{n+1}+C$

Thus,

$d(x)=\int v(x)dx=\int (2x-1)dx=x^2-x+C$

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Example Question #112 : Distance

The velocity of a particle is given by the function $v(t)=\frac{4}{cos^2(2t)}$ . Find the distance traveled by the particle over the interval of time t=[1,2] .

Possible Answers:

5.8

6.7

13.4

11.6

2.9

Correct answer:

6.7

Explanation:

Velocity is the time derivative of position, and by that token position can be found by integrating a known velocity function with respect to time:

$p(t)=\int v(t)dt$

Now, if this integral were to be taken over an interval of time t=[a,b] , this will give a finite value, a change in position, i.e. a distance travelled:

$d=\int_a^b v(t)dt$

For the velocity function

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

The distance travelled can be found via knowledge of the following derivative properties:

Trigonometric derivative:

$d[tan(u)]=\frac{du}{cos^2(u)}$

The distance travelled over t=[1,2] is:

d=2tan(2t)|_1^2

d=(2tan(2(2)))-2tan(2(1))=6.7

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Example Question #113 : How To Find Distance

The velocity of a particle is given by the function v(t)=3sin^2(t)cos(t) . Find the distance traveled by the particle over the interval of time $t=[0,\frac{\pi}{4}]$ .

Possible Answers:

$\frac{\sqrt{2}}{4}$

$\frac{\sqrt{2}}{8}$

$\frac{3\sqrt{2}}{4}$

$\frac{\sqrt{2}}{2}$

$\frac{3\sqrt{2}}{2}$

Correct answer:

$\frac{\sqrt{2}}{4}$

Explanation:

Velocity is the time derivative of position, and by that token position can be found by integrating a known velocity function with respect to time:

$p(t)=\int v(t)dt$

Now, if this integral were to be taken over an interval of time t=[a,b] , this will give a finite value, a change in position, i.e. a distance travelled:

$d=\int_a^b v(t)dt$

For the velocity function

v(t)=3sin^2(t)cos(t)

The distance travelled can be found via knowledge of the following derivative properties:

Trigonometric derivative:

d[sin(u)]=cos(u)du

w=sin(t);dw=cos(t)

$\int 3w^2dw=w^3=sin^3(t)$

The distance travelled over $t=[0,\frac{\pi}{4}]$ is:

$d=sin^3(t)|_0^{\frac{\pi}{4}}$

$d=sin^3(\frac{\pi}{4})-sin^3(0)=\frac{\sqrt{2}}{4}$

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Example Question #843 : Calculus

The velocity of a particle is given by the function v(t)=4t+8 . Find the distance traveled by the particle over the interval of time t=[1,3] .

Possible Answers:

Correct answer:

Explanation:

Velocity is the time derivative of position, and by that token position can be found by integrating a known velocity function with respect to time:

$p(t)=\int v(t)dt$

Now, if this integral were to be taken over an interval of time t=[a,b] , this will give a finite value, a change in position, i.e. a distance travelled:

$d=\int_a^b v(t)dt$

For the velocity function

v(t)=4t+8

The distance travelled over t=[1,3] is:

d=(2t^2+8t)|_1^3

d=(2(3)^2+8(3))-(2(1)^2+8(1))=32

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #101 : How To Find Distance

Example Question #101 : How To Find Distance

Example Question #102 : How To Find Distance

Example Question #103 : How To Find Distance

Example Question #102 : How To Find Distance

Example Question #841 : Calculus

Example Question #111 : Distance

Example Question #112 : Distance

Example Question #113 : How To Find Distance

Example Question #843 : Calculus

All Calculus 1 Resources

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