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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #2 : How To Find Position

The velocity of an object is given by the following equation:

v(t)=7t+4

If s(0)=5 , find the equation for the position of the object at any time .

Possible Answers:

s(t)=7t+9

s(t)=4t^2+7t+2

$s(t)=\frac{7}{2}t^2+4t+5$

s(t)=7t^2+2t+5

$s(t)=\frac{7}{2}t+5$

Correct answer:

$s(t)=\frac{7}{2}t^2+4t+5$

Explanation:

Velocity is the derivative of position, so in order to obtain an equation for position, we must integrate the given equation for velocity:

$s(t)=\int v(t)dt=\int (7t+4)dt=\frac{7}{2}t^2+4t+C$

The next step is to solve for C by applying the given initial condition, s(0)=5:

$s(t)=\frac{7}{2}t^2+4t+C$

$s(0)=0+0+C=5\rightarrow C=5$

So our final equation for position is:

$s(t)=\frac{7}{2}t^2+4t+5$

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Example Question #2 : How To Find Position

The position function of a ball from the ground when it is thrown by a pitcher is p(t)=4t+2 .

Where is the ball located at t=1 ?

Possible Answers:

p(1)=1

p(1)=4

p(1)=3

p(1)=6

p(1)=2

Correct answer:

p(1)=6

Explanation:

To find the position of the ball, we plug in t=1

So p(t)=4t+2 turns into:

$p(1)=4\cdot 1+2$

p(1)=4+2

p(1)=6

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Example Question #5 : How To Find Position

Function v(t) gives the velocity of a particle as a function of time.

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Find the equation that models the particle's postion as a function of time.

Possible Answers:

$\small \small h(t)=\frac{t^6}{5}+\frac{3t^5}{4}-\frac{7t^3}{9}+c$

$\small \small h(t)=\frac{t^6}{6}+\frac{3t^5}{7}-\frac{7t^3}{27}+c$

$\small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

$\small \small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}+\frac{7t^3}{9}+c$

$\small \small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}$

Correct answer:

$\small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

Explanation:

Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.

We are starting with the following

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

We need to perform the following:

$\small \small \small \small\int v(t)dt=\int t^5+3t^4-\frac{7t^2}{3}dt$

Recall that to integrate, we add one to each exponent and divide by the that number, so we get the following. Don't forget your +c as well.

$\small \small \int t^5+3t^4-\frac{7t^2}{3}dt=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

Which makes our position function, h(t), the following:

$\small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

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Example Question #3 : How To Find Position

Consider the velocity function modeled in meters per second by v(t).

v(t)=4t^3-5t^2+6

Find the position of a particle whose velocity is modeled by v(t) after seconds.

Possible Answers:

$393.3\bar{3}m$

$833.3\bar{3}+c m$

$8393.3\bar{3}+c \text{ }\frac{m}{s}$

$93.3\bar{3}+c {m}$

$8393.3\bar{3} m$

Correct answer:

$8393.3\bar{3}+c \text{ }\frac{m}{s}$

Explanation:

Recall that velocity is the first derivative of position, so to find the position function we need to integrate v(t) .

$V(t)=\int 4t^3-5t^2+6dt$

Becomes,

$V(t)=t^4-\frac{5t^3}{3}+6t+c$

Then, we need to find V(10)

$V(10)=10^4-\frac{5\cdot10^3}{3}+6\cdot10+c=8393.3\bar{3}+c$

So our final answer is:

$8393.3\bar{3}+c \text{ }\frac{m}{s}$

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Example Question #11 : How To Find Position

If the velocity of an object is represented by v(t)=30t^2 , what is the position of the object at t=2 ?

Possible Answers:

120

240

Correct answer:

Explanation:

To find the position given the velocity curve, take the antiderivative of v(t)=30t^2 .

$s(t)=\int v(t)dt=30\int t^2=30\left[\frac{t^{2+1}}{2+1} \right ]=30\left[\frac{t^3}{3} \right ]=\frac{30t^3}{3}=10t^3$

Solve for,

s(t=2) .

s(t=2)=10(2^3)=80

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Example Question #1 : Integration

Find the position function given the velocity function:

v(t)=3t^5+4t

Possible Answers:

$\frac{1}{2}t^6+\frac{1}{2}t^2$

15t^4+4

3t^6+4t^2

$\frac{1}{2}t^6+2t^2$

t^6+t^2

Correct answer:

$\frac{1}{2}t^6+2t^2$

Explanation:

To find the position from the velocity function, integrate

v(t)=3t^5+4t by increasing the exponent of each t term and then dividing that term by the new exponent value.

$s(t)=\int v(t)dt= \frac{3}{6}t^6+2t^2=\frac{1}{2}t^6+2t^2$

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Example Question #11 : Position

Suppose the velocity function of an object is v(t)=6t+2 . What is the position of the object at time t=2 ?

Possible Answers:

Correct answer:

Explanation:

To find the position function given the velocity, the velocity function needs to be integrated.

$s(t)=\int v(t)dt= \int (6t+2)dt$

s(t)= 3t^2+2t

Solve for s(t=2) .

s(t=2)=3(2)^2+2(2)= 12+4=16

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Example Question #12 : How To Find Position

Suppose an object's acceleration function is a(t)=2t-1 . The object's velocity is . What should be the object's position at that velocity?

Possible Answers:

$\frac{2}{3}$

There is not enough information.

$\frac{9}{2}$

Correct answer:

$\frac{9}{2}$

Explanation:

Integrate a(t)=2t-1 to obtain the velocity function, v(t) .

$v(t)=\int a(t)dt=\int (2t-1) dt=t^2-t$

Since velocity is 6, v(t)=6 . Substitute and solve for time.

6=t^2-t

0=t^2-t-6

0=(t-3)(t+2)

t=3,-2

Since negative time does not exist, the only valid solution for t at this velocity is t=3 .

Integrate the velocity function v(t) to find the position function, s(t) .

$s(t)=\int v(t)dt=\frac{t^3}{3}-\frac{t^2}{2}$

Substitute t=3 to s(t) to obtain the position.

$\frac{3^3}{3}-\frac{3^2}{2}= \frac{27}{3}-\frac{9}{2}=9-\frac{9}{2}=\frac{9}{2}$

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Example Question #863 : Spatial Calculus

Consider the velocity function given by v(t) :

v(t)=5t^4-343t^6+51t^2

Find the equation which models the position of a particle if its velocity can be modeled by v(t) .

Possible Answers:

p(t)=t^5-49t^7+17t^3

p(t)=60t^2-10290t^4+102

p(t)=120t-30870t^3

p(t)=t^5-49t^7+17t^3+c

p(t)=20t^3-2058t^5+102t

Correct answer:

p(t)=t^5-49t^7+17t^3+c

Explanation:

Recall that velocity is the first derivative of position and acceleration is the second derivative of position. Therfore, we need to integrate v(t) to find p(t)

v(t)=5t^4-343t^6+51t^2

$p(t)=V(t)=\int 5t^4-343t^6+51t^2dt$

So our answer is:

p(t)=t^5-49t^7+17t^3+c

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Example Question #861 : Spatial Calculus

Consider the velocity function given by v(t) :

v(t)=5t^4-343t^6+51t^2

Find the position of a particle after seconds if its velocity can be modeled by v(t) and the graph of its position function passes through the point (2,2) .

Possible Answers:

3828979

-38289.79

-3828.979

Correct answer:

Explanation:

Recall that velocity is the first derivative of position and acceleration is the second derivative of position. Therfore, we need to integrate v(t) to find p(t)

v(t)=5t^4-343t^6+51t^2

$p(t)=V(t)=\int 5t^4-343t^6+51t^2dt$

So we get:

p(t)=t^5-49t^7+17t^3+c

What we ultimately need is p(5), but first we need to find c: Use the point (2,2)

p(t)=t^5-49t^7+17t^3+c

$2=2^5-49\cdot 2^7+17\cdot 2^3+c$

c=6106

So our position function is:

p(t)=t^5-49t^7+17t^3+6106

$p(5)=5^5-49\cdot 5^7+17\cdot 5^3+6106=-3816769$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2 : How To Find Position

Example Question #2 : How To Find Position

Example Question #5 : How To Find Position

Example Question #3 : How To Find Position

Example Question #11 : How To Find Position

Example Question #1 : Integration

Example Question #11 : Position

Example Question #12 : How To Find Position

Example Question #863 : Spatial Calculus

Example Question #861 : Spatial Calculus

All Calculus 1 Resources

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