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Calculus 1 : Distance

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Example Questions

Example Question #21 : How To Find Distance

Let B=(2a-x,y) . Let A=(x,y) . What is the distance between these two points?

Possible Answers:

|x-y|

2|x+y|

$\sqrt{2}|x-a|$

2|a|

x+y

Correct answer:

$\sqrt{2}|x-a|$

Explanation:

Let A=(x,y) and (2a-x,y) .

We will use the distance formula to find the distance d(A,B).

By definition we have:

$D(A,B)=\sqrt{(2x-2a)^2+(y-y)^2}=|x-a|\sqrt{2}$

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Example Question #751 : Spatial Calculus

What is the average function value of y=x^2 from the interval [1,3] ?

Possible Answers:

$\frac{8}{3}$

$\frac{13}{3}$

Correct answer:

$\frac{13}{3}$

Explanation:

Write the formula for average function value.

$f_{average}= \frac{1}{b-a}\int_{a}^{b} f(x)dx$

Substitute the function and the bounds.

$f_{average}= \frac{1}{3-1}\int_{1}^{3}x^2dx$

$f_{average}= \left(\frac{1}{2}\right)\left[\frac{1}{3}x^3\right]_{1}^{3}=\left(\frac{1}{2}\right)\left[\frac{1}{3}(3)^3-\frac{1}{3}(1)^3\right]$

$f_{average}= \frac{1}{2}\left[9-\frac{1}{3}\right]=\frac{1}{2}\left[\frac{26}{3}\right]=\frac{13}{3}$

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Example Question #21 : Distance

An object is fired in the air, and the function which describes the object's height is y=-9x^2+8x+1 , where is in seconds. What is the height of the firing point to the maximum height of the object?

Possible Answers:

$\frac{25}{9}$

$\frac{16}{9}$

$\frac{5}{4}$

Correct answer:

$\frac{16}{9}$

Explanation:

Use the vertex formula to determine the location maximum height of the parabola.

$x=-\frac{b}{2a}=-\frac{8}{2(-9)}=\frac{4}{9}$

Substitute this value to the position function to find the height.

$y=-9\left(\frac{4}{9}\right)^2+8\left(\frac{4}{9}\right)+1$

$y=-9\left(\frac{16}{81}\right)+\frac{32}{9}+1$

$y=-\frac{16}{9}+\frac{32}{9}+\frac{9}{9}=\frac{25}{9}$

This is the maximum height, but at the start point x=0 , the y-intercept shows that the projectile is fired at a height of 1. Subtract the maximum from the height of the projectile.

$\frac{25}{9}-1=\frac{25}{9}-\frac{9}{9}= \frac{16}{9}$

The height from firing point to the maximum is $\frac{16}{9}$ .

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Example Question #22 : Distance

Find the midpoint of the line segment connecting the points (0,3,-4) and (2,-6,3) .

Possible Answers:

(1,2,1)

(0.5,-2,-1)

(2,-3,-1)

(1,-1.5,-0.5)

Correct answer:

(1,-1.5,-0.5)

Explanation:

The coordinates of the midpoint is the average of the coordinates of each point.

$\left(\frac{x_1+x_1}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

Substituting in the values given:

$\left(\frac{0+2}{2},\frac{3+(-6)}{2},\frac{-4+3}{2}\right)$

$\left(\frac{2}{2},\frac{-3}{2},\frac{-1}{2}\right)$

Therefore the coordinates of the midpoint are (1,-1.5,-0.5) .

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Example Question #21 : How To Find Distance

If the acceleration of an object is $f(t) = 2+\sin(t)$ . What is the displacement of the object from t=0 to t=2 , if the object had an initial velocity of ?

Possible Answers:

$\sin(2)$

$4 - \sin(2)$

$4+\sin(2)$

$8 - \sin(2)$

Correct answer:

$8 - \sin(2)$

Explanation:

The equation for displacement can be found by integrating the acceleration equation twice. Given the acceleration equation of $f(t)=2+\sin(t)$ .

The velocity equation is:

$v(t) = 2t - \cos(t) + C$

We can find the value of C using the initial velocity

v(0) = 0 -1 +C = 1

C = 2

The equation for velocity is then the integral of the acceleration function.

$v(t) = 2t - \cos(t) +2$

The equation of position is then

$x(t) = \int v(t)=\int 2t - \cos(t) +2$

$x(t) = t^2 -\sin(t) +2t + C$

Solving for the distance between t=2 and t=0, we solve for x(2) and x(0).

$x(2) = 4 - \sin(2)+4+C$

x(0) = 0 - 0 + 0 + C

We can now subtract x(0) from x(2) to find our distance

$x(2) - x(0) = 4 - \sin(2)+4+ C - 0 +0 - 0 - C$

$x(2) - x(0) = 4 - \sin(2)+ 4 = 8-\sin(2)$

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Example Question #21 : Distance

Find the midpoint of the line segment between the two points (5,7,0) and (-1,3,6) .

Possible Answers:

(4,10,6)

(2,5,3)

(5,4,6)

(-4,10,6)

Correct answer:

(2,5,3)

Explanation:

The midpoint can be found by taking the average of each of the coordinates.

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

Substituting in our values we find the midpoint as follows.

$\left(\frac{5-1}{2},\frac{7+3}{2},\frac{0+6}{2}\right) = (2,5,3)$

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Example Question #27 : How To Find Distance

Find the distance from (2,-4,6) to point (1,-3,9) .

Possible Answers:

$\sqrt{58}$

$\sqrt{11}$

$\sqrt{59}$

$\sqrt{15}$

Correct answer:

$\sqrt{11}$

Explanation:

Write the distance formula.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Substitute the point values and solve for distance.

$D=\sqrt{(1-2)^2+(-3-(-4))^2+(9-6)^2}=\sqrt{1+1+9}$

$D=\sqrt{11}$

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Example Question #22 : Distance

What is the midpoint of the line segment between the two points (8,3,2) and (10,-3,-4) ?

Possible Answers:

(-2,6,-6)

(12,0,5)

(9,0,-1)

(18,0,-2)

Correct answer:

(9,0,-1)

Explanation:

The midpoint is the average of the coordinates.

Therefore, to find the midpoint, we must add each coordinate of the first point to each coordinate of the second point and divide by two, finding the halfway point between the two points.

$\left(\frac{8+10}{2},\frac{3+(-3)}{2},\frac{2+(-4)}{2}\right) = (9,0,-1)$

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Example Question #23 : Distance

The velocity of a particle is given as v(t) = 8t -4 . What is the distance travelled by the particle t = 0 to t = 2 ?

Possible Answers:

Correct answer:

Explanation:

Given the velocity equation v(t) = 8t -4 , the position equation is the integral of the velocity from 0 to 2. To find this integral we can use the power rule.

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$

Therefore, the integral of the velocity equation is

$x(t) = \int_{0}^{2}(8t-4)dt = 4t^2-4t]_{0}^{2}$ .

To evaluate this integral from t = 0 to t = 2 , we now substitute in the value for when t = 2 and substract the values for when t = 0 .

4(2)^2 - 4(2) - 4(0)^2 + 4(0) = 16 - 8 - 0 + 0 = 8

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Example Question #763 : Spatial Calculus

The velocity of an object is given by the equation v(t) = 9t^2 - 4t^3 . What is the distance travelled by the object from t = 2 to t = 3 ?

Possible Answers:

Correct answer:

Explanation:

Given the velocity equation v(t) = 9t^2 - 4t^3 , we can solve for the position equation by taking the integral of the velocity.

To do this we must use the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Therefore, the integral of the velocity equation is

$x(t) = \int_{2}^{3}(9t^2-4t^3) = 3t^3 - t^4]_{2}^3$ .

We can now solve this by subtracting the value at t = 2 from the value when t = 3 .

$3t^3 - t^4]_{2}^3= 3(3)^3-(3)^4-3(2)^3+2^4 = 81 - 81 -24 +16 = -8$

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Calculus 1 : Distance

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #21 : How To Find Distance

Example Question #751 : Spatial Calculus

Example Question #21 : Distance

Example Question #22 : Distance

Example Question #21 : How To Find Distance

Example Question #21 : Distance

Example Question #27 : How To Find Distance

Example Question #22 : Distance

Example Question #23 : Distance

Example Question #763 : Spatial Calculus

All Calculus 1 Resources

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