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Calculus 1 : Position

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Example Questions

Example Question #941 : Spatial Calculus

What is the position function of the particle if the velocity is given by the following function:

v(t)=23t^2+e^t+20

Possible Answers:

$\frac{23t^3}{3}+te^t+20t+C$

$\frac{23t^3}{3}+e^t+20+C$

$\frac{23t^3}{3}+e^t+20t+C$

23t^3+e^t+20t+C

Correct answer:

$\frac{23t^3}{3}+e^t+20t+C$

Explanation:

The position function is equal to the integral of the velocity function:

$\int (23t^2+e^t+20)dt=\frac{23t^3}{3}+e^t+20t+C$

and was found using the following rules:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int e^xdx=e^x+C$

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Example Question #942 : Spatial Calculus

Find the position function s(t) if

v(t)=4t^2+15 .

Possible Answers:

$s(t)=\frac{t^3}{3}+15+c$

s(t)=4t^3+15t^2+c

$s(t)=\frac{4}{3}t^3+15t+c$

s(t)=4t^2+15+c

Correct answer:

$s(t)=\frac{4}{3}t^3+15t+c$

Explanation:

In order to find the position function from the velocity function we need to take the integral of the velocity function since

$s(t)=\int v(t)\,dt$ .

When taking the integral, we will use the inverse power rule which states,

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

Applying this rule to each term we get

$\int v(t)\,dt = \int 4t^2+15\,dt$

$= \frac{4t^3}{3}+15t+c$ .

As such,

$s(t)=\frac{4}{3}t^3+15t+c$ .

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Example Question #943 : Spatial Calculus

Find the position function if

$v(t)=cos \,t$ .

Possible Answers:

$s(t)= tan\, t + c$

$s(t)= cos\, t + c$

$s(t)= sec\, t + c$

$s(t)= sin\, t + c$

Correct answer:

$s(t)= sin\, t + c$

Explanation:

In order to find the position function from the velocity function we need to take the integral of the velocity function since

$s(t)=\int v(t)\,dt$ .

When taking the integral, we will use the trigonometric integral,

$\int cos\,x \, dx = sin\,x+c$ .

Applying this rule we get

$\int v(t)\,dt = \int cos\,t\,dt$

$=sin\,t+c$ .

As such,

$s(t)= sin\, t + c$ .

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Example Question #941 : Spatial Calculus

The acceleration of a particle is given by the function a(t)=6t+24 .

If the particle has an initial velocity of five and an initial position of ten, what will its position be at time t=3 ?

Possible Answers:

150

160

135

145

Correct answer:

160

Explanation:

Position can be found by integrating acceleration with respect to time twice (or integrating velocity with respect to time once):

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

For the acceleration function

a(t)=6t+24

First find velocity:

Use the following rule to find velocity,

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

v(t)=3t^2+24t+C_1

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

v(0)=5=3(0^2)+24(0)+C_1

C_1=5

v(t)=3t^2+24t+5

Now integrate once more to find position:

x(t)=t^3+12t^2+5t+C_2

To find this second contant, use the inital position:

x(0)=10=0^3+12(0^2)+5(0)+C_2

C_2=10

x(t)=t^3+12t^2+5t+10

x(3)=27+108+15+10

x(3)=160

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Example Question #945 : Spatial Calculus

A particle's velocity function is defined with respect to time as $v(t)=cos(\frac{1}{2}t)-sin(t)+2$ .

If it has an initial position of zero, what is its position at time $t=\pi$ ?

Possible Answers:

$2\pi+3$

$2(\pi+1)$

$2\pi$

$2(\pi-1)$

Correct answer:

$2\pi$

Explanation:

The position function can be found by integrating the velocity function with respect to time:

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

For the velocity function $v(t)=cos(\frac{1}{2}t)-sin(t)+2$ , the position function is therefore:

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

To find the constant of integration, use the initial position:

$x(0)=2sin(\frac{1}{2}0)+cos(0)+2(0)+C=0$

1+C=0

C=-1

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

$x(\pi)=2sin(\frac{1}{2}\pi)+cos(\pi)+2(\pi)-1$

$x(\pi)=2-1+2(\pi)-1$

$x(\pi)=2\pi$

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Example Question #91 : How To Find Position

A particle's velocity is given by the function: $v(t)=\frac{1}{2}te^{t^2}-sin(\frac{1}{3}t)$ . If the particle has an initial position of , what is its position at time t=2 ?

Possible Answers:

$\frac{1}{2}e^4-\frac{1}{6}sin\left(\frac{2}{3}\right)-4$

$\frac{1}{2}e^4-\frac{1}{6}sin\left(\frac{2}{3}\right)$

$e^3+3cos\left(\frac{2}{3}\right)+4$

$e^4+3cos\left(\frac{2}{3}\right)$

$\frac{1}{4}e^4+3cos\left(\frac{2}{3}\right)-\frac{13}{4}$

Correct answer:

$\frac{1}{4}e^4+3cos\left(\frac{2}{3}\right)-\frac{13}{4}$

Explanation:

The position function can be found by integrating the velocity function with respect to time:

$x(t)=\int v(t)=\int (\frac{1}{2}te^{t^2}-sin(\frac{1}{3}t))dt$

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

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

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

Thus we get the following position function.

$x(t)=\frac{1}{4}e^{t^2}+3cos\left(\frac{1}{3}t\right)+C$

The constant of integration can be found by using the initial condition:

$x(0)=\frac{1}{4}e^{0^2}+3cos\left(\frac{1}{3}(0)\right)+C=0$

$\frac{1}{4}+3+C=0$

$C=-\frac{13}{4}$

Knowing this provides the complete position function:

$x(t)=\frac{1}{4}e^{t^2}+3cos\left(\frac{1}{3}t\right)-\frac{13}{4}$

$x(2)=\frac{1}{4}e^4+3cos\left(\frac{2}{3}\right)-\frac{13}{4}$

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Example Question #111 : Calculus Review

A particle's velocity in two dimensions is described by the functions:

v_x(t)=t^2+sin(t)

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

If the particle has an initial position of (0,0) , what will its position be at time $t=2\pi$ ?

Possible Answers:

$(\frac{8\pi^3}{3},6\pi)$

$(\frac{8\pi^3-1}{3},6\pi-1)$

$(\frac{8\pi^3}{3}-1,6\pi)$

$(\frac{8\pi^3}{3}+1,6\pi-1)$

$(\frac{4\pi^3}{3},2\pi)$

Correct answer:

$(\frac{8\pi^3}{3},6\pi)$

Explanation:

Position can be found by integrating velocity with respect to time:

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

For velocities:

v_x(t)=t^2+sin(t)

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

The position functions are:

$x(t)=\frac{t^3}{3}-cos(t)+C_x$

$y(t)=\frac{sin(t^2)}{2}+3t+C_y$

These constants of integration can be found by using the given initial conditions:

$x(0)=\frac{0^3}{3}-cos(0)+C_x=0$

C_x=1

$y(0)=\frac{sin(0^2)}{2}+3(0)+C_y=0$

C_y=0

Which gives the definite integrals:

$x(t)=\frac{t^3}{3}-cos(t)+1$

$y(t)=\frac{sin(t^2)}{2}+3t$

$x(2\pi)=\frac{(2\pi)^3}{3}-cos(2\pi)+1$

$x(2\pi)=\frac{8\pi^3}{3}$

$y(2\pi)=\frac{sin((2\pi)^2)}{2}+3(2\pi)$

$y(2\pi)=6\pi$

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Example Question #93 : How To Find Position

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

a_x(t)=6t+4

a_y(t)=12t^2+2

The particle has the following initial velocities and positions:

$v_{0x}=2$

$v_{0y}=1$

x_0=3

y_0=4

What is the particle's position at time t=3 ?

Possible Answers:

(51,93)

(51,97)

(48,94)

(54,97)

(45,90)

Correct answer:

(54,97)

Explanation:

Position can be found by integrating with respect to time velocity once, or integrating with respect to time acceleration twice:

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

Begin by finding velocity functions. For the acceleration functions

a_x(t)=6t+4

a_y(t)=12t^2+2

Velocities are:

$v_x(t)=3t^2+4t+C_{1x}$

$v_y(t)=4t^3+2t+C_{1y}$

Find the constants of integration by using the given initial conditions:

$v_x(0)=3(0^2)+4(0)+C_{1x}=2$

$C_{1x}=2$

$v_y(0)=4(0^3)+2(0)+C_{1y}=1$

$C_{1y}=1$

Plugging these values back in gives the definite velocity functions:

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

v_y(t)=4t^3+2t+1

Now, find the position functions by integrating once more:

$x(t)=t^3+2t^2+2t+C_{2x}$

$y(t)=t^4+t^2+t+C_{2y}$

Find these new constants of integration the same way as before:

$x(0)=0^3+2(0^2)+2(0)+C_{2x}=3$

$C_{2x}=3$

$y(0)=0^4+0^2+0+C_{2y}=4$

$C_{2y}=4$

And the definite position functions are thus:

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

y(t)=t^4+t^2+t+4

Then at time t=3 , the particle's position is:

x(3)=3^3+2(3)^2+2(3)+3=27+18+6+3=54

y(3)=3^4+3^2+3+4=81+9+3+4=97

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Example Question #93 : How To Find Position

Find the position function x(t) of a race car at time t=4 if the velocity function is

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

Possible Answers:

$12.67\ meters$

$-6.45\ meters$

$37.33333\ meters$

$29.66667\ meters$

$35.2\ meters$

Correct answer:

$37.33333\ meters$

Explanation:

To find the position function for the race car, we must integrate the velocity function.

To integrate this function, we must use the following formula:

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

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

Now, to find the position at time t=4 , we plug in t=4 into the position function.

$p(4)=\frac{256}{4}-\frac{128}{3}+16 = 37\frac{1}{3} meters$

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Example Question #94 : How To Find Position

Tom rolls a ball up a hill at a velocity of 10 inches per second. It slows down at a rate of 4 inches per second. How many seconds will it take to come back to Tom?

Possible Answers:

2.5

4.5

Correct answer:

Explanation:

To find when it will get back to Tom. Assume Tom is at position 0, and find where the position function equals zero. Position funtion is the integral of the velocity funtion which is the intergral of the acceleration function.

a(t)=-4

$v(t)=\int-4dt=-4x+v(0)$

using,

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

v(0)=10 by what's given.

v(t)=-4t+10

$p(t)=\int(-4t+10)dt=-2t^2+10t+p(0)$

p(0)=0 since we assume Tom is at position 0, so now we must find where it is 0 again.

$-2t^2+10t=0 \Rightarrow -2t(t+5)=0 \Rightarrow t=0,5$

Thus, it will return to Tom at posistion after 5 seconds.

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Calculus 1 : Position

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #941 : Spatial Calculus

Example Question #942 : Spatial Calculus

Example Question #943 : Spatial Calculus

Example Question #941 : Spatial Calculus

Example Question #945 : Spatial Calculus

Example Question #91 : How To Find Position

Example Question #111 : Calculus Review

Example Question #93 : How To Find Position

Example Question #93 : How To Find Position

Example Question #94 : How To Find Position

All Calculus 1 Resources

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