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Calculus 1 : Calculus

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Example Questions

Example Question #791 : Spatial Calculus

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 #792 : Calculus

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 #793 : Calculus

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 #791 : Calculus

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:

150 meters

225 meters

125 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 #791 : Spatial Calculus

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 #796 : Calculus

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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Example Question #61 : Distance

A boat has a velocity defined by the equation v(t)=16t-10 . How much distance does it cover between t=1 and t=2 ?

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

$\int_{1}^{2}v(t)dt=\int_{1}^{2}(16t-10)dt=\left [ 8t^{2}-10t\right ]_{1}^{2}$

Completing the integral:

$\left [ 8t^{2}-10t\right ]_{1}^{2}=(8(2)^{2}-10(2))-(8(1)^{2}-10(1))=12+2=14$ .

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Example Question #61 : Distance

A particle starts at the origin and has velocity:

v(t)=t-3 .

How far does it travel in the first five seconds? (Hint: this question is not asking for the distance between the starting point and the endpoint, it is asking for the total distance traversed).

Possible Answers:

6.5

-2.5

2.5

4.5

-6.5

Correct answer:

6.5

Explanation:

First, we must will figure out the particle's position with respect to time by integrating v(t) with respect to . This gives us:

$p(t)=\frac{1}{2}t^2-3t+C$ .

We know that C=0 because we start at the origin. Thus,

$p(t)=\frac{1}{2}t^2-3t$ .

This is a quadratic equation, so we know it has one local extrema. We can solve for this by settings its derivative equal to 0:

v(t)=0

Solving this tells us that the extrema is at t=3 . Plugging this into p(t) tells us that the particle is at x=-4.5 when it turns around. So we know that it starts at x=0 moves to x=-4.5 and then turns around. After it turns around it goes to

$p(5)=\frac{1}{2}5^2-3\cdot5=-2.5$ .

This means it travels another (-2.5)-(-4.5)=2 units.

Therefore, its total distance traveled is

|-4.5|+|2|=6.5 units.

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Example Question #62 : Distance

Find the distance that the object has travelled in seconds given the initial velocity and acceleration.

v(0)=-14,a(t)=28

Possible Answers:

d(20)=2660

d(20)=2800

d(20)=5320m

d(20)=10920m

Correct answer:

d(20)=5320m

Explanation:

We integrate the acceleartion function and use the inital velocity to find the velocity function:

$\\ \int a(t)dt=v(t)=\int 28dt=28t+C\\ v(0)=-14=C\Rightarrow v(t)=28t-14$

We can now either choose to find the position function or take the definite integral of the velocity function over the given time interval to find the distance directly. Since we are not given the initial position, this is the natural approach.

$\\ \int_0^{20}v(t)dt=d(t)\\ \\ \int_0^{20}28t-14dt=14t^2-14t]_0^{20}=14\cdot400-14\cdot20=5320=d(t)$

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Example Question #62 : Distance

A ball is thrown upwards at a speed of 10m/s from a 100 m building. Assume gravity is 9.8m/s .

What is the approximate total distance the ball has travelled (up and down) from its point of release to its impact with the ground?

Possible Answers:

90m

115m

110 m

105 m

100 m

Correct answer:

110 m

Explanation:

The max height of the ball happens when the velocity is zero. Using our equation,

$v(t) = \int a(t) dt = \int -9.8 dt = -9.8t+v_0=-9.8t+10$ we can solve for the value of for which the velocity is zero.

This occurs at approximately t=1 . When t=1 , the height of the ball is 105 m .

This means the ball travels meters up from the throw at the top of the building and then drops 105 m to the ground, for a total of 110m travelled.

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #791 : Spatial Calculus

Example Question #792 : Calculus

Example Question #793 : Calculus

Example Question #791 : Calculus

Example Question #791 : Spatial Calculus

Example Question #796 : Calculus

Example Question #61 : Distance

Example Question #61 : Distance

Example Question #62 : Distance

Example Question #62 : Distance

All Calculus 1 Resources

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