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Calculus 1 : Distance

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Example Questions

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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Example Question #31 : Distance

A boat has a velocity defined by the equation v(t)=10t+3 . How far did the boat travel between t=1 and t=2 ?

Possible Answers:

Correct answer:

Explanation:

In order to find the distance traveled by the boat from t=1 to t=2 we need to integrate the velocity function:

$\int_{1}^{2}v(t)dt=\int_{1}^{2}(10t+3)dt=\left [5t^{2}+3t \right ]_{1}^{2}$

Solving the integral,

$\left [5t^{2}+3t \right ]_{1}^{2}=(5(2)^{2}+3(2))-(5(1)^{2}+3(1))$

=(20+6)-(5+3)=26-8=18

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Example Question #771 : Calculus

A particle has a velocity defined by the equation v(t)=40t+5 . How far did the particle travel between t=3 and t=4 ?

Possible Answers:

145

164

166

168

163

Correct answer:

145

Explanation:

In order to find the distance traveled by the particle from t=3 to t=4 we need to integrate the velocity function.

\ $\int_{3}^{4}v(t)dt=\int_{3}^{4}(40t+5)dt=\left [20t^{2}+5t \right ]_{3}^{4}$

Solving the integral,

$\left [20t^{2}+5t \right ]_{3}^{4}=(20(4)^{2}+5(4))-(20(3)^{2}+5(3))$

=(320+20)-(180+15)=340-195=145

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Example Question #37 : Distance

The velocity equation of an object is given by the equation v(t) = 15t . What is the distance covered by the object from t = 0 to t = 1 ?

Possible Answers:

7.5

225

Correct answer:

7.5

Explanation:

The distance travelled by the object is equal to the intergral of the velocity equation from t = 0 to t = 1 .

To take the integral of the velocity equation we can use the power rule which says that if:

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

Applying this to our function we get:

$x(t) = \int_0^1 (15t)dt = \frac{15t^{1+1}}{1+1}]_0^1 = \frac{15t^{2}}{2}]_0^1$

$x(1) - x(0) = \frac{15(1)^2}{2} - \frac{15(0)^2}{2} = \frac{15}{2} - 0 = 7.5$

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Example Question #31 : Distance

The velocity of an object is given by the equation $v(t) = \cos(t)$ . What is the distance covered by the object from t= 0 to t = 2 if the object has an initial velocity of

Possible Answers:

$\sin(2) - 1$

$-\sin(2)$

$\sin(2)$

$\sin(2)+1$

Correct answer:

$\sin(2)$

Explanation:

The distance can be found by integrating the velocity of the object from t = 0 to t = 2 .

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

Therefore,

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

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Example Question #39 : Distance

The acceleration of an object is given by the equation $a(t) = 100 - \sin(t)$ . What is the distance travelled by the object from t = -1 to t = 1 , if the initial velocity of the object is ?

Possible Answers:

100

$18 +\sin(1) - \sin(-1)$

Correct answer:

$18 +\sin(1) - \sin(-1)$

Explanation:

To find the distance travelled by the object we must first integrate the acceleration equation to find the velocity equation. The integration can be done using the power rule where if

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

We can rewrite te acceleration equation as $a(t) = 100t^0 - \sin(t)$ and apply this rule to integrate the equation.

$v(t) = \int (100t^0 - \sin(t))dt = \frac{100t^{0+1}}{0+1} + \cos(t) +C = 100t + \cos(t) +C$

We can solve for the value of by using the initial velocity of the object

$v(0) = 100(0) + \cos(0) +C =1+ C= 10$ .

Therefore C = 9 and $v(t) = 100t +\cos(t) + 9$

$x = \int_{-1}^1 (100t^1 +\cos(t) + 9t^0) dt$

$= \frac{100t^{1+1}}{1+1} +\sin(t)+ \frac{9t^{0+1}}{0+1}]_{-1}^1$

$=50t^2 +\sin(t)+9t]_{-1}^1$

We can now evaluate this as

$50t^2 +\sin(t) +9t]_{-1}^1 = 50(1)^2 +\sin(1) +9(1) - 50(-1)^2-\sin(-1)-9(-1)$

$= 50 + \sin(1) + 9 - 50 -\sin(-1) +9 = 18 + \sin(1) -\sin(-1)$

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Example Question #771 : Calculus

The acceleration of an object is given by the equation a(t) = 14 . What is the distance travelled by the object between time t= 0 and t =1 , if the initial velocity of the object is ?

Possible Answers:

Correct answer:

Explanation:

The distance travelled by the object can be found by integrating the acceleration equation to find velocity, and then integrating the velocity equation from t= 0 to t = 1 . To integrate the acceleration equation we must use the power rule where if

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

Therefore the velocity equation of the object is

$v(t) = \int (14)dt = \frac{14t^{(0+1)}}{0+1} +C = 14t +C$

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

v(0) = 14(0) + C =0 + C= 3

Therefore C =3 and v(t) = 14t + 3

We now integrate the velocity equation from t = 0 to t =1 to find the distance travelled by the object

$s = \int_0^1 (14t+3)dt = \frac{14t^{(1+1)}}{1+1} +\frac{3t^{(0+1)}}{0+1}]_0^1 = 7t^2 + 3t]_0^1$

s = 7(1)^2 + 3(1) - 7(0)^2 + 3(0) = 7 + 3 - 0 + 0 = 10

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Calculus 1 : Distance

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #31 : How To Find Distance

Example Question #764 : Spatial Calculus

Example Question #32 : Distance

Example Question #765 : Spatial Calculus

Example Question #31 : Distance

Example Question #771 : Calculus

Example Question #37 : Distance

Example Question #31 : Distance

Example Question #39 : Distance

Example Question #771 : Calculus

All Calculus 1 Resources

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