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Calculus 1 : Distance

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Example Question #51 : Distance

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 #52 : Distance

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 #53 : Distance

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 #54 : 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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Example Question #55 : Distance

A given object has a velocity defined by the equation v(t)=8t-5 . How far does it travel between t=1 and t=2 ?

Possible Answers:

Correct answer:

Explanation:

In this example, we are looking to find the distance the object has traveled between t=1 and t=2 .

Distance can be defined as the definite integral of velocity. For this particular problem we will use the power rule which states,

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

Therefore, we need to find

$\int_{1}^{2}v(t)dt=\int_{1}^{2}(8t-5)dt=\left [ 4t^{2}-5t\right ]_{1}^{2}$ .

Solving this integral, we get:

$\left [ 4t^{2}-5t\right ]_{1}^{2}$

$=[4(2)^{2}-5(2)]-[4(1)^{2}-5(1)]$

=[16-10]-[4-5]

=6-[-1]

=6+1

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Example Question #56 : Distance

A car has a velocity defined by the equation v(t)=6t+7 . How far does it travel between t=3 and t=4 ?

Possible Answers:

Correct answer:

Explanation:

In this example, we are looking to find the distance the car has traveled between t=3 and t=4 .

Distance can be defined as the definite integral of velocity. For this particular problem we will use the power rule which states,

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

Therefore, we need to find

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

Solving this integral, we get:

$\\ \left [ 3t^{2}+7t\right ]_{3}^{4}=(3(4)^{2}+7(4))-(3(3)^{2}+7(3)) \\ \\=(48+28)-(27+21)\\ \\=76-48\\ \\=28$

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Example Question #57 : Distance

A hummingbird has a velocity defined by the equation v(t)=10t+1 . How far does it travel between t=4 and t=5 ?

Possible Answers:

Correct answer:

Explanation:

In this example, we are looking to find the distance the hummingbird has traveled between t=4 and t=5 .

Distance can be defined as the definite integral of velocity.

For this problem we will use the power rule when integrating which states,

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

Therefore, we need to find

$\int_{4}^{5}v(t)dt=\int_{4}^{5}(10t+1)dt=\left [ 5t^{2}+t\right ]_{4}^{5}$ .

Solving this integral, we get:

$\left [ 5t^{2}+t\right ]_{4}^{5}$

$=[5(5)^{2}+(5)]-[5(4)^{2}+(4)]$

=[125+5]-[80+4]

=130-84

=46

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Example Question #58 : Distance

A race car is traveling at a constant 50 m/s when the driver suddenly hits the brakes. Assuming a constant deceleration of 10 m/s², how far will the car travel before it comes to a complete stop?

Possible Answers:

125 meters

150 meters

225 meters

Not enough information is given.

Correct answer:

125 meters

Explanation:

To find the change in position of the car, let us start with the car's acceleration as a function of time:

a(t)

Since acceleration doesn't change with time, it has a constant value.

$a(t) = -10 m/s^{2}$

(the negative is used to represent deceleration).

We can then integrate this function with respect to time to find velocity.

$v(t) = ( -10 m/s^2 ) \cdot t + v_{0}$

Where v₀ represents the initial velocity, which was given to us as 50 m/s.

We want to know the time where the car comes to rest, meaning where v(t_r) = 0.

Solving

$(-10 m/s^2 )\cdot t_{r} + 50 m/s = 0$

for t_rgives our time at rest, 5 s.

Now, we can integrate our velocity function with respect to time to find our position function.

$x(t_{r}) = x_{0} + v_{0}t_{r} + (1/2)at_{r}^2$

Since we're not interested in the absolute position at the time of rest, but rather the change in position, we can move the x₀ term to the other side of the equation:

$x(t_{r}) - x_{0} = v_{0}t_{r} + (1/2)at_{r}^2$

Plugging in our values for

$v_{0}:(50 m/s)$

a:(-10m/s^2)

and

t_r:(5 s)

We can find how far the car travelled after the brakes were hit, our

$(x(t_{r}) - x_0)$ quantity, to be 125 m .

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Example Question #59 : Distance

A cheetah has a velocity defined by the equation v(t)=18t-4 . How much distance does it cover between t=4 and t=5 ?

Possible Answers:

Correct answer:

Explanation:

In order to find the distance travelled between t=4 and t=5 , we need to take the definite integral of the velocity v(t) .

Let's first define the definite integral as

$\int_{a}^{b}f(t)dt=F(b)-F(a)$

for a continuous function over a closed interval [a,b] with an antiderivative .

Using the inverse of the power rule

$\int t^{n}dt=\frac{t^{n+1}}{n+1}+C$

with a constant and $n\neq-1$ , we can therefore determine that

$\int_{4}^{5}v(t)dt=\int_{4}^{5}(18t-4)dt=\left [ 9t^{2}-4t\right ]_{4}^{5}$ .

Completing the integral:

$\left [ 9t^{2}-4t\right ]_{4}^{5}=(9(5)^{2}-4(5))-(9(4)^{2}-4(4))=205-128=77$

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Example Question #51 : How To Find Distance

A train has a velocity defined by the equation v(t)=12t+7 . How much distance does it cover between t=3 and t=4 ?

Possible Answers:

Correct answer:

Explanation:

In order to find the distance travelled between t=3 and t=4 , we need to take the definite integral of the velocity v(t) .

Let's first define the definite integral as

$\int_{a}^{b}f(t)dt=F(b)-F(a)$ for a continuous function over a closed interval [a,b] with an antiderivative .

Using the inverse of the power rule

$\int t^{n}dt=\frac{t^{n+1}}{n+1}+C$

with a constant and $n\neq-1$ , we can therefore determine that

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

Completing the integral:

$\left [ 6t^{2}+7t\right ]_{3}^{4}=(6(4)^{2}+7(4))-(6(3)^{2}+7(3))=124-75=49$ .

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Calculus 1 : Distance

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #51 : Distance

Example Question #52 : Distance

Example Question #53 : Distance

Example Question #54 : Distance

Example Question #55 : Distance

Example Question #56 : Distance

Example Question #57 : Distance

Example Question #58 : Distance

Example Question #59 : Distance

Example Question #51 : How To Find Distance

All Calculus 1 Resources

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