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Calculus 1 : Calculus

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Example Questions

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 #761 : Spatial Calculus

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

Possible Answers:

(4,10,6)

(5,4,6)

(2,5,3)

(-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 #762 : Spatial Calculus

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

Possible Answers:

(18,0,-2)

(9,0,-1)

(12,0,5)

(-2,6,-6)

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 #764 : Calculus

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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Example Question #31 : How To Find Distance

The velocity of an object is given by the equation v(t) = 12t -6t^2 . What is the distance travelled by the object from t = 0 to t = 1 ?

Possible Answers:

Correct answer:

Explanation:

The distance travelled can be found by integrating the velocity equation v(t) = 12t -6t^2 .

The velocity equation is integrated by using the following rule.

$g'(x) = x^n \Rightarrow g(x) = \frac{x^{n+1}}{n+1}$

Applying this rule to v(t) gives,

$x(t) = \int_{0}^{1}(12t - 6t^2) dt =\frac{12}{1+1}t^{(1+1)} - \frac{6}{2+1}t^{(2+1)} = 6t^2 - 2t^3]_{0}^{1}$ .

The the distance is now calculated by subtracting the position at t = 0 from the position at t = 1 .

$6t^2 - 2t^3]_{0}^{1} = 6(1)^2 - 2(1)^3-0+0 = 6-2 = 4$

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Example Question #764 : Spatial Calculus

The acceleration of an object is a(t) = t^4 . What is the approximate distance covered by the object from t= 2 to t = 3 if the object has an initial velocity of ?

Possible Answers:

10.7

22.2

44.6

110.7

Correct answer:

22.2

Explanation:

The distance of the object can be found by differentiating the acceleration equation a(t) = t^4 twice. To differentiate the acceleration equation we can use the power rule where if

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

Appying this rule to the acceleration equation gives us,

$v(t) = \int (t^4)dt = \frac{t^{4+1}}{4+1} +C = \frac{t^5}{5} +C$ .

We can find the value of by using the initial velocity of the object.

$v(0) = \frac{0^5}{5} +C = 0+C = 0$

Therefore, C = 0 and $v(t) = \frac{t^5}{5}$ .

We can now find the distance covered by the object by integrating the velocity equation from to .

$x(t) = \int_2^3 \left(\frac{t^{5}}{5}\right)dt = \left(\frac{t^{5+1}}{5(5+1)}\right)|_2^3 = \frac{t^6}{5(6)}|_2^3 = \frac{t^6}{30}|_2^3$

Evaluating this equation gives

$x(3)-x(2) = \frac{(3)^6}{30} - \frac{(2)^6}{30} = \frac{3(243)}{3(10)}= \frac{243}{10} - \frac{64}{30} \approx 24.3 - 2.1 \approx 22.2$

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Example Question #32 : Distance

The position of an object is given by the equation x(t)=4t^2 - t +3 . What is the distance between the position of the object at time t = 0 and time t =3 ?

Possible Answers:

34.5

Correct answer:

Explanation:

To solve for the distance, we can use the position equation given to us to find the location of the object at t= 0 and t= 3 . The distance is the difference between these to locations.

x(3) - x(0) = 4(3)^2 - (3) + 3 - 4(0)^2 + (0) - 3 = 36 - 0 - 0 + 0 -3 =33

Therefore the distance from the location of the object at t = 0 to the location at t= 3 is

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Example Question #765 : Spatial Calculus

A car has a velocity defined by the equation v(t)=20t-4 . How far did the car travel between t=4 and t=5 ?

Possible Answers:

Correct answer:

Explanation:

In order to find the distance traveled by the car from t=4 to t=5 we need to set up the integral of the velocity function: $\int_{4}^{5}v(t)dt=\int_{4}^{5}(20t-4)dt=\left [10t^{2}-4t \right ]_{4}^{5}$

Solving the integral,

$\left [10t^{2}-4t \right ]_{4}^{5}=(10(5)^{2}-4(5))-(10(4)^{2}-4(4))$

=(250-20)-(160-16)=230-144=86

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #21 : How To Find Distance

Example Question #761 : Spatial Calculus

Example Question #27 : How To Find Distance

Example Question #762 : Spatial Calculus

Example Question #764 : Calculus

Example Question #763 : Spatial Calculus

Example Question #31 : How To Find Distance

Example Question #764 : Spatial Calculus

Example Question #32 : Distance

Example Question #765 : Spatial Calculus

All Calculus 1 Resources

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