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Calculus 1 : Distance

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Example Questions

Example Question #111 : Distance

Find the distance function given the following velocity function:

v(x)=2x-1

Possible Answers:

d(x)=0

d(x)=x^2-x

d(x)=2

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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Example Question #113 : Distance

The velocity of a particle is given by the function v(t)=8sin(t)cos(t) . Find the distance traveled by the particle over the interval of time $t=[0,\frac{\pi}{2}]$ .

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)=8sin(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)

$\int8wdw=4w^2$

The distance travelled over $t=[0,\frac{\pi}{2}]$ is:

$d=4sin^2(t)|_0^{\frac{\pi}{2}}$

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

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Example Question #112 : Distance

Find the distance from points: (-3,-4) to (5,-5)

Possible Answers:

$\sqrt{106}$

$\sqrt{65}$

Correct answer:

$\sqrt{65}$

Explanation:

This is simply the application of the distance formula:

The distance is going to be equal to:

$d=\sqrt{(y_1-y_0)^2+(x_1-x_0)^2}=\sqrt{(-4+5)^2+(5+3)^2}=\sqrt{65}$

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Example Question #115 : Distance

Determine the distance travelled by a person if their displacement is 100m , and they moved equal lengths West and North, and didn't move in any other direction.

Possible Answers:

$50\sqrt{2}m$

100m

$100\sqrt{2}m$

$200\sqrt{2}m$

Correct answer:

$100\sqrt{2}m$

Explanation:

We know that displacement is just the magnitude of the vector moving from point to point . Distance however is the sum of the movements. In this question, we can treat the distance as the hypotenuse of a triangle, and the movements in the west and north directions as the 2 legs. Since the person moved the same distance in each direction, which I will call , we can determine it by doing:

100^2= a^2+a^2

$\sqrt{\frac{100^2}{2}}=50\sqrt{2}=a$

Since the person walked in both directions, the total distance travelled is:

$50\sqrt{2}+50\sqrt{2}=100\sqrt{2}=a$

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Example Question #113 : Distance

Determined the displacement of a person who walks 30m west and 70m north.

Possible Answers:

$10\sqrt{58}m$

100m

-40m

$20\sqrt{10}$

Correct answer:

$10\sqrt{58}m$

Explanation:

The displacement in orthogonal directions(North and West) can be determined by using the pythagorean theorem:

$d=\sqrt{30^2+70^2}=\sqrt{5800}=10\sqrt{58}$

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Calculus 1 : Distance

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #111 : Distance

Example Question #112 : Distance

Example Question #113 : How To Find Distance

Example Question #843 : Calculus

Example Question #113 : Distance

Example Question #112 : Distance

Example Question #115 : Distance

Example Question #113 : Distance

All Calculus 1 Resources

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