Calculus 1 : Distance

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #771 : Spatial Calculus

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

Possible Answers:

\(\displaystyle 27\)

None of these.

\(\displaystyle 7\)

\(\displaystyle 14\)

Correct answer:

\(\displaystyle 14\)

Explanation:

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

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

Using this rule, the distance is calculated as

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

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

 

Example Question #771 : Calculus

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

Possible Answers:

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

\(\displaystyle x =1\)

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

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

Correct answer:

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

Explanation:

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

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

Therefore

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

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

Example Question #771 : Spatial Calculus

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

Possible Answers:

\(\displaystyle 0\)

\(\displaystyle -8\)

\(\displaystyle 4\)

\(\displaystyle -11\)

Correct answer:

\(\displaystyle -11\)

Explanation:

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

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

Using the power rule the distance travelled by the object is

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

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

Example Question #42 : How To Find Distance

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

Possible Answers:

\(\displaystyle -3\)

\(\displaystyle 0\)

\(\displaystyle 4\)

\(\displaystyle 3\)

Correct answer:

\(\displaystyle -3\)

Explanation:

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

The velocity can be integrated using the power rule where

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

Applying this to the velocity equation gives

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

Example Question #41 : Distance

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

Possible Answers:

\(\displaystyle 47\)

\(\displaystyle 43\)

\(\displaystyle 55\)

\(\displaystyle 46\)

Correct answer:

\(\displaystyle 46\)

Explanation:

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

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

Therefore the distance between the locations is \(\displaystyle 46\).

Example Question #43 : How To Find Distance

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

Possible Answers:

\(\displaystyle 6,450\)

\(\displaystyle -30,000\)

\(\displaystyle 70,000\)

\(\displaystyle 30,000\)

Correct answer:

\(\displaystyle -30,000\)

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

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

Using this equation we find that, 

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

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

\(\displaystyle s = 15,000 - 50,000 +5,000 = -30,000\).

Example Question #781 : Calculus

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

Possible Answers:

None of these.

\(\displaystyle 24\)

\(\displaystyle 32\)

\(\displaystyle 8\)

Correct answer:

\(\displaystyle 32\)

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

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

Therefore the distance covered by the object is

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

Example Question #45 : How To Find Distance

The acceleration of an object is given by the equation \(\displaystyle a(t) = 6t\). What is the distance covered by the object from time \(\displaystyle t =0\) to \(\displaystyle t =1\), if the initial velocity of the object is \(\displaystyle 3\)

Possible Answers:

\(\displaystyle 4\)

\(\displaystyle 2\)

\(\displaystyle 7\)

\(\displaystyle 3\)

Correct answer:

\(\displaystyle 4\)

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

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

Using this rule gives 

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

The value of \(\displaystyle C\) can be found using the initial velocity of the object.

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

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

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

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

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

Example Question #781 : Spatial Calculus

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

Possible Answers:

\(\displaystyle 5\)

\(\displaystyle 4\)

\(\displaystyle 6\)

\(\displaystyle 14\)

Correct answer:

\(\displaystyle 6\)

Explanation:

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

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

Applying this to the velocity equation gives

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

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

Example Question #45 : Distance

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

Possible Answers:

\(\displaystyle 41\)

\(\displaystyle 22\)

\(\displaystyle 23\)

\(\displaystyle 32\)

Correct answer:

\(\displaystyle 23\)

Explanation:

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

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

Therefore the distance covered is

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

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

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