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Calculus 1 : Velocity

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Example Questions

Example Question #201 : Calculus

The position of a particle in two dimensions is given by the functions x(t)=t^2+3t-4 , y(t)=3x^2(t) . What is the particle's velocity in the (x,y) directions at time t=3 ?

Possible Answers:

(9,81)

(3,243)

(9,756)

(6,588)

(9,486)

Correct answer:

(9,756)

Explanation:

Velocity is the time derivative of position:

$v(t)=\frac{d}{dt}p(t)$

In the case of two position functions for the description of two-dimensional location, there will simply be two corresponding velocity functions:

For the position functions

x(t)=t^2+3t-4

y(t)=3x^2(t)

The velocity functions are then:

v_x(t)=2t+3

$v_y(t)=6x(t)\frac{dx(t)}{dt}$

v_y(t)=6x(t)v_x(t)

v_y(t)=6(t^2+3t-4)(2t+3)

At time t=3

v_x(t)=6+3=9

v_y(t)=6(9+9-4)(6+3)=6(14)(9)=756

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Example Question #202 : Velocity

The position of a particle in three-dimensional space is given by the functions

$x(t)=3cos(\frac{1}{2}t)$

y(t)=2sin(t)cos(t)

z(t)=t^2+t

What is the magnitude of the particle's velocity at time $t=\pi$ ?

Possible Answers:

$\sqrt{8\pi+\frac{29}{4}}$

$\sqrt[3]{2\pi+\frac{3}{2}}$

$\sqrt[3]{4\pi^2+4\pi+\frac{29}{4}}$

$\sqrt{4\pi^2+4\pi+\frac{29}{4}}$

$\sqrt{2\pi+\frac{3}{2}}$

Correct answer:

$\sqrt{4\pi^2+4\pi+\frac{29}{4}}$

Explanation:

Velocity can be found as the time derivative of position:

$v(t)=\frac{d}{dt}p(t)$

In the case of three-dimensional descriptions, that simply means there are three velocity functions. For the positional functions

$x(t)=3cos(\frac{1}{2}t)$

y(t)=2sin(t)cos(t)

z(t)=t^2+t

The resultant velocity functions are:

$v_x(t)=-\frac{3}{2}sin(\frac{1}{2}t)$

v_y(t)=2cos^2(t)-2sin^2(t)

v_z(t)=2t+1

At time $t=\pi$

$v_x(\pi)=-\frac{3}{2}sin(\frac{1}{2}\pi)=-\frac{3}{2}$

$v_y(\pi)=2cos^2(\pi)-2sin^2(\pi)=2$

$v_z(\pi)=2\pi+1$

The magnitude of the velocity is then the root of the sum of the squares:

$v(\pi)=\sqrt{(-\frac{3}{2})^2+(2)^2+(2\pi+1)^2}$

$v(\pi)=\sqrt{\frac{9}{4}+4+4\pi^2+4\pi+1}$

$v(\pi)=\sqrt{4\pi^2+4\pi+\frac{29}{4}}$

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Example Question #202 : Calculus

The acceleration of a particle is given by the function $a(t)=9t^2+15e^{-3t}$ . If the particle has an initial velocity of , what is it's velocity after five seconds?

Possible Answers:

$360+5e^{-15}$

$390-15e^{-15}$

$360-5e^{-15}$

$390-5e^{-15}$

$375-5e^{-15}$

Correct answer:

$390-5e^{-15}$

Explanation:

Velocity can be found by integrating acceleration with respect to time:

$v(t)=\int a(t)dt$

To find the integral we will need to apply a few different rules.

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

$\int e^u=e^u\cdot \frac{1}{\frac{du}{dx}}$

For the acceleration function

$a(t)=9t^2+15e^{-3t}$ ,

the velocity function is thus:

$v(t)=3t^3-5e^{-3t}+C$

To find the constant of integration, use the initial velocity condition:

$v(0)=3(0^3)-5e^{-3(0)}+C=10$

C=15

From here, the full velocity function can be found:

$v(t)=3t^3-5e^{-3t}+15$

$v(5)=3(5^3)-5e^{-3(5)}+15$

$v(5)=390-5e^{-15}$

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Example Question #204 : Velocity

The position of a particle in three dimensions is given by the functions:

x(t)=2t+4

y(t)=x^3(t)

z(t)=2y^2(t)

What is the velocity at time t=2 ?

Possible Answers:

(32,384,1384681)

(4,5260,984213)

(2,298,842352)

(2,716,432)

(2,384,786432)

Correct answer:

(2,384,786432)

Explanation:

Velocity can be found as the time derivative of position:

$v(t)=\frac{d}{dt}p(t)$

For the three position functions

x(t)=2t+4

y(t)=x^3(t)

z(t)=2y^2(t)

There can be found three velocity funcitons:

v_x(t)=2

$v_y(t)=3x^2(t)\frac{dx(t)}{dt}$

$v_z(t)=4y(t)\frac{dy(t)}{dt}$

Now note that these derivative terms correspond to velocities, i.e. $v_x(t)=\frac{dx(t)}{dt}$ and $v_y(t)=\frac{dy(t)}{dt}$ .

Knowing this, we can replace terms in each function so that each velocity is in terms of time:

v_x(t)=2

v_y(t)=3(2t+4)^2(2)

v_z(t)=4(2t+4)^3[3(2t+4)^2(2)]

Therefore at time t=2

v_x(2)=2

v_y(2)=3(4+4)^2(2)=384

v_z(2)=4(4+4)^3(384)=786432

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Example Question #205 : Velocity

The position of a particle in two dimensions is given by the functions:

$x(t)=4cos(\frac{1}{2}t)+sin(3t)$

$y(t)=2sin(\frac{1}{2}t)cos(\frac{3}{2}t)$

What is the magnitude of the particle's velocity at time $t=\pi$ ?

Possible Answers:

$\sqrt2$

$\sqrt{34}$

$2\sqrt{2}$

Correct answer:

$\sqrt{34}$

Explanation:

Velocity can be found as the time derivative of position:

$v(t)=\frac{d}{dt}p(t)$

For the two position functions

$x(t)=4cos(\frac{1}{2}t)+sin(3t)$

$y(t)=2sin(\frac{1}{2}t)cos(\frac{3}{2}t)$

The velocity functions are found to be:

$v_x(t)=-2sin(\frac{1}{2}t)+3cos(3t)$

$v_y(t)=cos(\frac{1}{2}t)cos(\frac{3}{2}t)-3sin(\frac{1}{2}t)sin(\frac{3}{2}t)$

The velocities at time $t=\pi$ are found to be:

$v_x(\pi)=-2sin(\frac{1}{2}\pi)+3cos(3\pi)=-2-3=-5$

$v_y(\pi)=cos(\frac{1}{2}\pi)cos(\frac{3}{2}\pi)-3sin(\frac{1}{2}\pi)sin(\frac{3}{2}\pi)$

$v_y(\pi)=0(0)-3(1)(-1)=3$

The magnitude of the velocity can be found by taking the root of the sum of squares:

$v(\pi)=\sqrt{(-5)^2+(3)^2}$

$v(\pi)=\sqrt{34}$

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Example Question #202 : Spatial Calculus

The acceleration of a football is given by $a(t)=3t^{2}+2t$ . What is the velocity function v(t) of the football if the initial velocity is $3 \frac{meters}{sec}$ ?

Possible Answers:

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

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

$v(t)=3t^{3}+2t$

$v(t)=t^{3}+t^{2}+3$

$v(t)=t^{3}+4t^{2}-3$

Correct answer:

$v(t)=t^{3}+t^{2}+3$

Explanation:

To find the velocity function v(t) , we need to integrate the acceleration function a(t) .

To integrate this function, we need the following formulae:

$\int x^{n}dx=\frac{1}{n+1}x^{n+1}, \int cf(x)dx=c\int f(x)dx$

$v(t)=\int a(t) = \int 3t^{2}+2t = \frac{3t^{3}}{3}+\frac{2t^2}{2}+C=t^{3}+t^{2}+C$

Now, to solve for the constant, we use the initial conditions:

Since v(0)=3 , we can plug in for in the velocity function and set it equal to :

$v(0)=3(0)^{2}+2(0)+C=0+0+C=3$

Therefore, C=3

So, the specific solution for the velocity is $v(t)=t^{3}+t^{2}+3$ .

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Example Question #203 : Spatial Calculus

A particle's velocity is given by the function v(t)=3t^2+4t+6 . What is the average velocity of the particle over the interval of time t=[2,10] ?

Possible Answers:

157.5

154

123.2

126

Correct answer:

154

Explanation:

Average velocity can be found as the total distance travelled over the total time expended.

Distance travelled can be found by integrating the velocity function over the specified interval of time t=[a,b] .

Therefore

$v_{avg}=\frac{1}{b-a}\int_a^b v(t)dt$

For the velocity function

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

$v_{avg}=\frac{1}{10-2}\int_2^{10} (3t^2+4t+6)dt$

$v_{avg}=\frac{1}{8}(t^3+2t^2+6t)|_2^{10}$

$v_{avg}=\frac{1}{8}[(1000+200+60)-(8+8+12)]$

$v_{avg}=\frac{1}{8}[(1260)-(28)]$

$v_{avg}=\frac{1}{8}(1232)$

$v_{avg}=154$

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Example Question #203 : Calculus

The velocity of a particle is given by the function v(t)=t^2+t+1 . What is the particle's average velocity over the period of time t=[0,3] ?

Possible Answers:

16.5

5.5

Correct answer:

5.5

Explanation:

Average velocity can be found by dividing total distance traveled over total time elapsed.

When given a velocity function, distance traveled over an interval of time t=[a,b] can be found by integrating the velocity over this interval. By dividing by the length of this interval, an average velocity can be found:

$v_{avg}=\frac{1}{b-a}\int_a^b v(t)dt$

For the velocity function

v(t)=t^2+t+1

The average velocity over the interval t=[0,3] can be found as

$v_{avg}=\frac{1}{3-0}\int_0^3(t^2+t+1)dt$

$v_{avg}=\frac{1}{3}(\frac{t^3}{3}+\frac{t^2}{2}+t)|_0^3$

$v_{avg}=\frac{1}{3}[(9+\frac{9}{2}+3)-(0+0+0)]$

$v_{avg}=\frac{11}{2}=5.5$

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Example Question #204 : Calculus

The position of a particle is given by the function x(t)=t^2-10t+24 . At what time does the particle change directions in its movement?

Possible Answers:

Correct answer:

Explanation:

The particle changes directions when its velocity undergoes a sign change. Velocity can be found as the time derivative of position, so for the position function

x(t)=t^2-10t+24

The velocity can be found as

v(t)=2t-10

The particle initially travels in the negative direction, since v(0)=-10 . The velocity will change in sign after the point that the function reaches zero, if such a point exists for a positive time value:

2t-10=0

2t=10

t=5

This indicates the turning point. All later times, the particle will move in a positive direction.

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Example Question #205 : Calculus

A particle's position is given by the function x(t)=-t^3+9t^2+5t+20 . What is the maximum velocity achieved by the particle?

Possible Answers:

Correct answer:

Explanation:

Velocity can be found by taking the derivative of position with respect to time. For the position function

x(t)=-t^3+9t^2+5t+20

The velocity function is

v(t)=-3t^2+18t+5

Now to find where the velocity has minima or maxima, take the derivative again with respect to time to find acceleration, and find when the accerlation is zero:

a(t)=-6t+18

-6t+18=0

t=3

Prior to this time, the acceleration is postive, and after this time the acceleration is negative, so this indicates a maximum.

The maximum velocity is then

$v_{max}=v(3)=-27+54+5$

v_max=32

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Calculus 1 : Velocity

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #201 : Calculus

Example Question #202 : Velocity

Example Question #202 : Calculus

Example Question #204 : Velocity

Example Question #205 : Velocity

Example Question #202 : Spatial Calculus

Example Question #203 : Spatial Calculus

Example Question #203 : Calculus

Example Question #204 : Calculus

Example Question #205 : Calculus

All Calculus 1 Resources

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