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Calculus 1 : Spatial Calculus

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Example Question #41 : Calculus

Consider the position function {}p(t) , which describes the positon of an oxygen molecule.

p(t)=14t^3-14t^2+14t-14

Find the function which models the velocity of the oxygen molecule.

Possible Answers:

v(t)=84t-28

v(t)=-28t+14

v(t)=42t^2-28t+14

v(t)=42t^2+14

v(t)=14

Correct answer:

v(t)=42t^2-28t+14

Explanation:

Recall that velocity is the first derivative of position and acceleration is the second derivative of position.

So given:

p(t)=14t^3-14t^2+14t-14

Apply the power rule to each term to find the velocity.

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

v(t)=p'(t)=42t^2-28t+14

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Example Question #42 : Calculus

Consider the position function {}p(t) , which describes the positon of an oxygen molecule.

p(t)=14t^3-14t^2+14t-14

Find the velocity of the oxygen molecule after seconds.

Possible Answers:

57.4

2590

574

Correct answer:

574

Explanation:

Recall that velocity is the first derivative of position and acceleration is the second derivative of position.

So given:

p(t)=14t^3-14t^2+14t-14

v(t)=p'(t)=42t^2-28t+14

To find the velocity after 4 seconds, find v(4)

$v(4)=42\cdot 4^2-28\cdot 4+14=574$

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Example Question #43 : Calculus

The displacement of an object at time is defined by the function $f(t) = e^{(3t)}$ . What is the velocity at t=1 ?

Possible Answers:

Correct answer:

Explanation:

The velocity is defined as the derivative of the displacement equation. With the displacement equation $f(t) = e^{3t}$ , we can use the power rule to differentiate.

Remember that $f(x)=e^x \rightarrow f'(x)=e^x \cdot {dx}$ .

In our case our $x=3t \rightarrow \frac{dx}{dt}=3$

Thus, the velocity equation is defined as:

$f'(t) = 3e^{3t}$ . Now t=1 can be used in the velocity equation.

$f'(1) = 3e^{3} \approx 3 \cdot 20 = 60$

The velocity at t=1 is 60.

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Example Question #44 : Calculus

The position of an object is given by the equation $f(t) = \sin^2(x)$ . What is the velocity of the object?

Possible Answers:

$\cos^2(t)$

$\sin(t)$

$2\sin(t)$

$2\sin(t)\cos(t)$

Correct answer:

$2\sin(t)\cos(t)$

Explanation:

The velocity of the object can be found by differentiating the equation $f(t) = \sin^2(t)$ . We must use the chain rule and the power rule to differentiate this equation.

We can rewrite the equation as

f(t)=(sin(t))^2 using the power rule we get:

f'(t)=2sin(t) from here we need to use the chain rule and multiply this by the derivative of the inside piece which is the sine function. Since the derivative of sine is cosine we get the following:

$f'(t)=2sin(t)\cdot cos(t)$

Thus, the velocity equation is:

$f'(t) = 2\sin(t)\cos(t)$

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Example Question #41 : How To Find Velocity

The position of an object is given by the equation $f(t) = 2\sin(t)$ . What is the velocity of the object at t=0 ?

Possible Answers:

Correct answer:

Explanation:

The velocity equation can be found by differentiating the equation for position $f(t) = 2\sin(t)$ .

This gives the equation

$f'(t) = 2\cdot \cos(t)$

At t=0

$f'(0) = 2\cos(0) = 2\cdot 1 = 2$

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Example Question #46 : Calculus

The acceleration of an object is given by the equation f'(t) = t^2 - t^3 . What is the velocity equation of the object if it has an initial velocity of ?

Possible Answers:

$f(t) = \frac{t^3}{3}-\frac{t^4}{4}+2$

$f(t) = \frac{t^3}{3}-\frac{t^4}{4}$

f(t) = 2t - 3t^2

f(t) = 6

Correct answer:

$f(t) = \frac{t^3}{3}-\frac{t^4}{4}+2$

Explanation:

The velocity equation can be found by integrating the acceleration equation.

Differentiating the equation f'(t) = t^2 - t^3 gives

$f(t) = \frac{t^3}{3}- \frac{t^4}{4}+C$

Using the initial velocity of the equation gives us

f(0) = 0 - 0 +C = 2

C = 2

The complete equation for the velocity is

$f(t) = \frac{t^3}{3} - \frac{t^4}{4} +2$

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Example Question #41 : How To Find Velocity

If the acceleration of an object is $-3 \frac{m}{s}$ , and its acceleration function is a(t)=4t-6 , what is the velocity of the object?

Possible Answers:

$-\frac{27}{8}$

$\frac{17}{54}$

$-\frac{63}{8}$

Correct answer:

$-\frac{27}{8}$

Explanation:

To find the velocity from the acceleration function, integrate the acceleration function a(t)=4t-6 . To integrate, remember to add an exponent to each term and divide the whole term by the value of the exponent.

v(t)= 2t^2-6t

We will also need to know the time when the object at $-3 \frac{m}{s}$ .

Substitute this value back into the acceleration function and solve for time.

-3=4t-6

3=4t

$t=\frac{3}{4}$

Substitute this value to the velocity function to find the velocity.

$v\left(t=\frac{3}{4}\right)=2\left(\frac{3}{4}\right)^2-6\left(\frac{3}{4}\right)=2\left(\frac{9}{16}\right)-\frac{18}{4}=\frac{9}{8}-\frac{36}{8}$

$v=-\frac{27}{8}$

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Example Question #48 : Calculus

If the position function of a ball is s(t)=3t^2+4t+5 , what is the velocity of the ball when t=0 ?

Possible Answers:

Correct answer:

Explanation:

To find the velocity function, take the derivative of the position function: s(t)=3t^2+4t+5

Remember when integrating, increase the exponent of each term by one and then divide the term by the value of the new exponent.

v(t)=6t+4

Substitute the time into the velocity function.

v(t=0)=6(0)+4=4

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Example Question #49 : Calculus

The acceleration of an object is given by the equation f(t) = -2t . What is the velocity of the object at t=10 if its initial velocity is ?

Possible Answers:

150

2400

2600

{}-50

Correct answer:

{}-50

Explanation:

Using the acceleration equation, f'(t) = -2t we can find the velocity equation by taking the integral.

$f(t) = \frac{-2t^{1+1}}{2} + C = -t^2 +C\cdot$

Now using the initial velocity of 50, we can find

f(0) = -(0)^2+C = 50

C = 50

Now we can solve for the velocity at t=10

f(10) = -(10)^2+50 = -50

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Example Question #50 : Calculus

The position equation for an object is $f(t) = \sin^2(\pi e^t)$ . What is the velocity of the object at t = 0 ?

Possible Answers:

$2 \pi$

$-2 \pi$

Correct answer:

Explanation:

The velocity of the object can be found by differentiating the position equation. To differentiate this equation, the power rule and the chain rule must be used.

The power rule says that

$f(x) = x^{n} \rightarrow f'(x) = nx^{n-1}$ .

The chain rule says that

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\times g'(x)$ .

Differentiating the equation $f(t) = \sin^2(\pi e^t)$ , gives

$f'(t) = 2\sin^{(2-1)}(\pi e^t)\cos(\pi e^t)\pi e^t =2\sin(\pi e^t)\cos(\pi e^t)\pi e^t = 2\pi e^t\sin(\pi e^t)\cos(\pi e^t)$ .

Using 0 as the value for t gives us

$f'(0) = 2\pi e^0\sin(\pi e^0)\cos(\pi e^0) = 2\cdot\pi\cdot \sin(\pi)\cdot\cos(\pi) = 0$

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Calculus 1 : Spatial Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #41 : Calculus

Example Question #42 : Calculus

Example Question #43 : Calculus

Example Question #44 : Calculus

Example Question #41 : How To Find Velocity

Example Question #46 : Calculus

Example Question #41 : How To Find Velocity

Example Question #48 : Calculus

Example Question #49 : Calculus

Example Question #50 : Calculus

All Calculus 1 Resources

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