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Calculus 1 : Calculus

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Example Questions

Example Question #41 : Distance

The position of an object is given by the equation x(t) = 2t^3 - 3t +4 . What is the distance from the object's location at t = 1 to the object's location at t =3 ?

Possible Answers:

Correct answer:

Explanation:

To find the distance between the two locations we can subtract the position at t = 1 from the position at t = 3 .

$\\x(3) - x(1) = 2(3)^3 - 3(3) +4 - 2(1)^3 + 3(1) - 4 \\= 54 - 9 +4 -2 +3 -4 \\= 46$

Therefore the distance between the locations is .

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Example Question #42 : Distance

The velocity of an object is given by the equation v(t) = 300t - 20t^3 +500 . What is the distance covered by the object from t = 0 to t = 10 ?

Possible Answers:

6,450

-30,000

30,000

70,000

Correct answer:

-30,000

Explanation:

To find the distance travelled we can integrate the velocity equation of the object.

This can be done using the power rule where if

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

Using this equation we find that,

$s = \int_0^{10} (300t-20t^3+500t^0)dt = \frac{300t^{1+1}}{1+1} - \frac{20t^{3+1}}{3+1} +\frac{500t^{0+1}}{0+1}]_0^{10}$

s = 150(10)^2 - 5(10)^4 + 500(10) -150(0)^2 - 5(0)^4 +500(0)

s = 15,000 - 50,000 +5,000 = -30,000 .

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Example Question #43 : Distance

The velocity of an object is given by the equation v(t) = 8t^3 . What is the distance covered by the object from t = 0 to t = 2 ?

Possible Answers:

None of these.

Correct answer:

Explanation:

To find the distance covered by the object we can integrate the velocity equation. This can be done using the power rule where if

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

Therefore the distance covered by the object is

$s = \int_0^2 (8t^3)dt = \frac{8t^{(3+1)}}{3+1}]_0^2 = 2t^4]_0^2 = 2(2)^4 - 2(0)^4 = 32$ .

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Example Question #781 : Calculus

The acceleration of an object is given by the equation a(t) = 6t . What is the distance covered by the object from time t =0 to t =1 , if the initial velocity of the object is ?

Possible Answers:

Correct answer:

Explanation:

The distance covered by the object can be found by integrating the acceleration twice. This can be done using the power rule where if

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

Using this rule gives

$v(t) = \int (6t)dt = \frac{6t^{(1+1)}}{1+1} + C = 3t^2 +C$ .

The value of can be found using the initial velocity of the object.

v(0) = 3(0)^2 + C = 0 +C = 3

Therefore C = 3 and v(t) = 3t^2 + 3 .

Integrating the velocity equation from t= 0 to t= 1 will give us the distance covered by the object.

$s= \int_0^1 (3t^2+3)dt = \frac{3t^{(2+1)}}{2+1} + \frac{3t^{(0+1)}}{0+1}]_0^1 = t^3 + 3t]_0^1$

s = (1)^3 + 3(1) - (0)^3 - 3(0) = 1 + 3 -0 - 0= 4

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Example Question #44 : Distance

The velocity of an object is given by the equation v(t) = 30t^2 - 4 . What is the distance covered by the object between t= 0 and t =1 ?

Possible Answers:

Correct answer:

Explanation:

The distance covered by the object can be found by integrating the velocity from t = 0 to t =1 , using the power rule where if

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

Applying this to the velocity equation gives

$s = \int_0^1 (30t^2-4)dt = \frac{30t^{2+1}}{2+1} - \frac{4t^{0+1}}{0+1}]_0^1 = 10t^3 - 4t]_0^1$

s = 10(1)^3 - 4(1) - 10(0)^3 +4(0) = 10 -4 - 0 + 0 = 6

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Example Question #45 : Distance

The velocity equation of an object is given by the equation v(t) = 20t - 3t^2 . What is the distance covered by the object from t= 1 to t =2 ?

Possible Answers:

Correct answer:

Explanation:

The distance covered can be found by integrating the velocity from t= 1 to t =2 using the power rule, where if

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

Therefore the distance covered is

$s = \int_1^2 (20t - 3t^2)dt = \frac{20t^{(1+1)}}{1+1} - \frac{3t^{2+1}}{2+1}]_1^2 = 10t^2 - t^3]_1^2$

s= 10(2)^2 - (2)^3 - 10(1)^2 + (1)^3 = 40 - 8 - 10 + 1 = 23

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Example Question #781 : Spatial Calculus

The velocity of an object is given by the equation v(t) = 10 - 12t^2 . What is the distance covered by the object from t =2 to t =3 ?

Possible Answers:

Correct answer:

Explanation:

The distance covered by the object is equal to the velocity of the object integrated from t =2 to t =3 . We can integrate the velocity equation using the power rule where if

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

The distance covered by the object is then

$s = \int_2^3 (10 -12t^2)dt = \frac{10t^{(0+1)}}{0+1} - \frac{12t^{2+1}}{2+1}]_2^3 = 10t - 4t^3]_2^3$

s = 10(3) - 4(3)^3 - 10(2) + 4(2)^3 = 30 - 108 - 20 +32 = -66

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Example Question #782 : Spatial Calculus

The velocity of a speedboat is defined by the equation v(t)=12t+7 . What distance does the speedboat travel between t=3 and t=4 ?

Possible Answers:

Correct answer:

Explanation:

The distance d(t) of an object traveled over a certain amount of time is equal to the definite integral of its velocity over that time.

For this particular problem we will use the power rule when integrating. The power rule states,

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

Therefore, in this instance,

$d(t)=\int_{3}^{4}v(t)dt=\int_{3}^{4}(12t+7)dt=\left [ 6t^{2}+7t\right ]_{3}^{4}$ .

Solving the integral, we get $[6(4)^{2}+7(4)]-[6(3)^{2}+7(3)]=124-75=49$ .

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Example Question #783 : Spatial Calculus

The velocity of a particle is defined by the equation v(t)=4t+6 . What distance does the particle travel between t=5 and t=6 ?

Possible Answers:

Correct answer:

Explanation:

The distance d(t) of an object traveled over a certain amount of time is equal to the definite integral of its velocity over that time.

For this particular problem we will use the power rule when integrating. The power rule states,

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

Therefore, in this instance,

$d(t)=\int_{5}^{6}v(t)dt=\int_{5}^{6}(4t+6)dt=\left [ 2t^{2}+6t\right ]_{5}^{6}$ .

Solving the integral, we get

$[2(6)^{2}+6(6)]-[2(5)^{2}+6(5)]=108-80=28$ .

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Example Question #52 : Distance

The velocity of a car is defined by the equation v(t)=2t+3 . What distance does the car travel between t=2 and t=3 ?

Possible Answers:

Correct answer:

Explanation:

The distance d(t) of an object traveled over a certain amount of time is equal to the definite integral of its velocity over that time.

For this particular problem we will use the power rule when integrating. The power rule states,

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

Therefore, in this instance,

$d(t)=\int_{2}^{3}v(t)dt=\int_{2}^{3}(2t+3)dt=\left [ t^{2}+3t\right ]_{2}^{3}$ .

Solving the integral, we get

$[(3)^{2}+3(3)]-[(2)^{2}+3(2)]=18-10=8$ .

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #41 : Distance

Example Question #42 : Distance

Example Question #43 : Distance

Example Question #781 : Calculus

Example Question #44 : Distance

Example Question #45 : Distance

Example Question #781 : Spatial Calculus

Example Question #782 : Spatial Calculus

Example Question #783 : Spatial Calculus

Example Question #52 : Distance

All Calculus 1 Resources

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