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Calculus 1 : Distance

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Example Questions

Example Question #772 : Calculus

The velocity of an object is given by the equation v(t) = 3t^2 - 5 . What is the distance covered by the object from time t = 2 to time t = 3 ?

Possible Answers:

None of these.

Correct answer:

Explanation:

The distance covered by the object can be found by integrating the velocity equation from time t= 2 to time t = 3 . To integrate the velocity equation we can use the power rule where if

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

Using this rule, the distance is calculated as

$x = \int_2^3 (3t^2 - 5)dt = \frac{3t^{(2+1)}}{2+1} - \frac{5t^{(0+1)}}{0+1}]_2^3 = t^3 - 5t]_2^3$

x = (3)^3 - 5(3) - (2)^3 + 5(2) = 27 - 15 - 8 +10 = 14

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Example Question #773 : Calculus

The velocity of an object is given by the equation $v(t) = \frac{1}{2}\cos(t)$ . What is the distance travelled by the object from time t= 0 to time t = 1 ?

Possible Answers:

$x = \frac{\sin(1)}{2}$

x =1

$x = \frac{\sin(1)}{2} - \frac{1}{2}$

$x = \frac{\sin(1)}{2} - \frac{1}{4}$

Correct answer:

$x = \frac{\sin(1)}{2}$

Explanation:

The distance travelled by the object can be found by integrating the velocity equation.

Because the derivative of $\sin(t)$ is $\cos(t)$ , the integral of $\cos(t)$ is $\sin(t)$ .

Therefore

$x = \int_0^1 (\frac{1}{2}\cos(t))dt = \frac{1}{2}\sin(t)]_0^1 = \frac{1}{2}\sin(1) - \frac{1}{2}\sin(0) = \frac{\sin(1)}{2} - 0$

$x = \frac{\sin(1)}{2}$

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Example Question #771 : Spatial Calculus

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

Possible Answers:

Correct answer:

Explanation:

The distance travelled by the object can be found by integrating the position equation for the object from t =1 to t =2 . This can be done using the power rule where if

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

Using the power rule the distance travelled by the object is

$s = \int_1^2 (2t - 6t^2)dt = \frac{2t^{(1+1)}}{1+1} - \frac{6t^{(2+1)}}{2+1}]_1^2 = t^2 - 2t^3]_0^1$ .

s = (2)^2 - 2(2)^3 - (1)^2 + 2(1)^3= 4 - 16 - 1 + 2 = -11

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Example Question #774 : Spatial Calculus

The velocity of an object is v(t) = 4t -7 . What is the distance travelled by the object from t= 0 to t = 3 ?

Possible Answers:

Correct answer:

Explanation:

The distance travelled can be found by integrating the velocity from t= 0 to t =3 .

The velocity can be integrated using the power rule where

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

Applying this to the velocity equation gives

$\\s = \int_0^3 (4t -7t^0)dt = \frac{4t^{(1+1)}}{1+1} -\frac{7t^{(0+1)}}{0+1}]_0^3 = 2t^2 - 7t]_0^3 \Rightarrow \\ \\ \Rightarrow 2(3)^2 - 7(3) - 2(0)^2 + 7(0) = 18 - 21 - 0 + 0 = -3$

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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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Calculus 1 : Distance

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #772 : Calculus

Example Question #773 : Calculus

Example Question #771 : Spatial Calculus

Example Question #774 : Spatial Calculus

Example Question #41 : Distance

Example Question #42 : Distance

Example Question #43 : Distance

Example Question #781 : Calculus

Example Question #44 : Distance

Example Question #45 : Distance

All Calculus 1 Resources

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