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Calculus 1 : Velocity

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Example Questions

Example Question #71 : How To Find Velocity

The position equation of an object is given by the equation $x(t) = \cos^3(e^{4t})$ . What is the velocity of the object?

Possible Answers:

$v(t) = -3e^{4t}\cos(e^{4t})\sin(e^{4t})$

$v(t) = -12e^{4t}\cos^2(e^{4t})\sin(e^{4t})$

$v(t) = -3e^{4t}\cos^2(e^{4t})\sin(e^{4t})$

$v(t) = -12\cos^2(e^{4t})\sin(e^{4t})$

Correct answer:

$v(t) = -12e^{4t}\cos^2(e^{4t})\sin(e^{4t})$

Explanation:

The velocity of the object can be found by differentiating the object's position. This equation can be accurately differentiated using the chain rule and the power rule.

The chain rule is,

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

and the power rule is,

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

Therefore the velocity of the object is

$v(t) = 3\cos^2(e^{4t})(-\sin(e^{4t}))(e^{4t})(4) = -12e^{4t}\cos^2(e^{4t})\sin(e^{4t})$ .

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Example Question #71 : How To Find Velocity

A particles position, , as a function of time is given by

$s(t)=\frac{1}{2}a_1t^2+a_2t+a_3$

where a_1, a_2, a_3 are constants.

Find equations for the velocity, , and acceleration, , of the particle.

Possible Answers:

v=a_1t^2+a_2

a=0

v=a_2

a=a_3t

v=a_1t+a_2

a=a_1

v=a_2t+a_3

a=0

v=a_1t^2

a=a_1

Correct answer:

v=a_1t+a_2

a=a_1

Explanation:

The velocity of a moving object is given by $\frac{ds}{dt}$ .

The acceleration of a moving object is given by $\frac{d^2s}{dt^2}$ .

We will need to differntiate once to get velocity and another time to get the acceleration. We will need to use the power rule when differentiating. The power rule is as follows.

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

Using these formulas we find that

$v=\frac{1}{2}\cdot 2a_1t^{2-1}+a_2t^{1-1}=a_1t+a_2$

$a=a_1t^{1-1}=a_1$

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Example Question #71 : How To Find Velocity

The position, , of a partice is given by the following equation.

s(t)=12t^3-6t^2-10

What is this particle's velocity at t=2 ?

Possible Answers:

120

104

112

Correct answer:

120

Explanation:

To find the velocity of the particle we first need to differentiate the position function. To take the derivative we will use the power rule.

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

Thus the velocity of a particle is found to be,

$v=\frac{ds}{dt}=36t^2-12t$ .

Plugging in for t=2, we find

v=120 .

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Example Question #71 : How To Find Velocity

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

Possible Answers:

$v(t) = -2t\cos(e^{t^2})$

$v(t) = 2\cos(e^{t^2})$

$v(t) = 2te^{t^2}\cos(e^{t^2})$

$v(t) = 2t\cos(e^{t^2})$

Correct answer:

$v(t) = 2te^{t^2}\cos(e^{t^2})$

Explanation:

The velocity of the object can be found by differentiating the position of the object. To accurately differentiate the position we can use the power rule and the chain rule.

The power rule is, $f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

The chain rule is, $f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x) \times g'(x)$ .

Differenting the object's position using these two equations gives

$v(t) = \cos(e^{t^2})(e^{t^2})((2)t^{(2-1)}) = 2te^{t^2}\cos(e^{t^2})$ .

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Example Question #75 : How To Find Velocity

The position of an object is given by the equation $x(t) = 10t^2\sin(t)$ . What is the velocity of the object?

Possible Answers:

$v(t) = 20t\sin(t) - 10t^2\cos(t)$

$v(t) = 20t\sin^2(t) - 10t^2\cos(t)$

$v(t) = 20t\sin(t) + 10t^2\cos(t)$

$v(t) = 20t\sin(t) - 10t\cos(t)$

Correct answer:

$v(t) = 20t\sin(t) + 10t^2\cos(t)$

Explanation:

The velocity of the object can be found by differentiating the object's position. To accurately differentiate the object's position we can use the product rule and the power rule where if

$f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$

and if

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

Using these rules the velocity equation is

$v(t) = 10(2)t^{(2-1)}\sin(t) + 10t^2\cos(t) = 20t\sin(t) + 10t^2\cos(t)$ .

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Example Question #71 : How To Find Velocity

The position of an object is given by the equation $x(t) = \cos^4(t)$ . What is the velocity of the object at t = 2 ?

Possible Answers:

$a(t) = -4\cos^3(2)\sin(2)$

$a(t) = -\cos^4(2)\sin(2)$

$a(t) = 4\cos^3(2)\sin(2)$

$a(t) = 2\cos^3(2)\sin(2)$

Correct answer:

$a(t) = -4\cos^3(2)\sin(2)$

Explanation:

The velocity of the object is equal to the derivative of the object's position. The velocity can be found using the power rule and the chain rule where if $f(x) = x^n \rightarrow f'(x) = nx^{n-1}$

and if

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

Therefore the velocity of the object is

$a(t) =(4)\cos^{(4-1)}(t)(-\sin(t)) = -4\cos^3(t)\sin(t)$ .

At time t =2 , the velocity is

$a(t) = -4\cos^3(2)\sin(2)$ .

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Example Question #71 : How To Find Velocity

The acceleration of an object is given by the equation $a(t) = \frac{-2}{t^3}$ . What is the velocity of the object at time t=1 , if the initial velocity of the object is ?

Possible Answers:

None of the above.

Correct answer:

Explanation:

The velocity of the object can be found by integrating the acceleration. This can be done using the power rule where if

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

Using this rule the velocity of the object is

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

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

$v(0) = (0)^{-2}+ C = 0 +C = 2$

Therefore C = 2 and $v(t) = t^{-2} + 2$ .

With the velocity equation we solve for the velocity at t= 1 .

$v(1) = (1)^{-2} + 2 = 1 + 2 = 3$

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Example Question #73 : How To Find Velocity

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

Possible Answers:

$v(t) = 2t\sin(t) - t^2\cos(t)$

$v(t) = 2\sin(t) - t^2\cos(t)$

$v(t) = 2t\sin(t) + t^2\cos(t)$

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

Correct answer:

$v(t) = 2t\sin(t) + t^2\cos(t)$

Explanation:

To find the velocity we can differentiate the position of the object. To do this accurately we use the product rule and the power rule.

The product rule is

$f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$ .

The power rule is

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

The velocity of the equation is then found to be

$v(t) = (2)t^{(2-1)}\sin(t) + t^2(\cos(t)) = 2t\sin(t) + t^2\cos(t)$ .

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Example Question #71 : How To Find Velocity

The position of an object is given by the equation x(t) = 4t^2 . What is the velocity of the object at time t =10 ?

Possible Answers:

400

1333

Correct answer:

Explanation:

The velocity of the object is equal to the derivative of the position. To differentiate the position we use the power rule where if

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

With this rule we find the velocity equation to be

$v(t) = x'(t) = 4(2)t^{(2-1)} = 8t$ .

Using time t = 10 , we can obtain a velocity of

v(10) = 8(10) = 80 .

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Example Question #74 : How To Find Velocity

The acceleration of an object is given by the equation a(t) = 12 . What is the velocity of the object, if the initial velocity of the object is ?

Possible Answers:

v(t) = 2t^3

v(t) = 6t^2

v(t) = 12t

v(t) = 12t - 12

Correct answer:

v(t) = 12t

Explanation:

To find the velocity we must integrate the acceleration using the power rule.

The power rule is where

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

Applying this to the acceleration equation gives

$v(t) = \int (12t^0)dt = \frac{12t^{(0+1)}}{0+1} + C = 12t +C$ .

Using the initial velocity we find the value of .

v(0) = 12(0) + C = 0 + C = 0

Therefore C = 0 and v(t) = 12t .

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Calculus 1 : Velocity

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #71 : How To Find Velocity

Example Question #71 : How To Find Velocity

Example Question #71 : How To Find Velocity

Example Question #71 : How To Find Velocity

Example Question #75 : How To Find Velocity

Example Question #71 : How To Find Velocity

Example Question #71 : How To Find Velocity

Example Question #73 : How To Find Velocity

Example Question #71 : How To Find Velocity

Example Question #74 : How To Find Velocity

All Calculus 1 Resources

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