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Calculus 1 : Spatial Calculus

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Example Questions

Example Question #941 : Spatial Calculus

Find the position function of a rocket shot from the ground if its initial velocity is $v_{0}=100ft/s$ and its acceleration is $a=-32ft/s^{2}$ .

Possible Answers:

$s(t)=100t-16t^{2}$

$s(t)=t^{2}+5t-12$

s(t)=-32t+100

s(t)=100

$s(t)=100t-32t^{2}$

Correct answer:

$s(t)=100t-16t^{2}$

Explanation:

The position function of an object moving with uniform acceleration is $s(t)=s_{0}+v_{0}t+\frac{1}{2}at^{2}$ , where $s_{0}$ is the inital position of the object, $v_{0}$ is the inital velocity of the object and is the acceleration of the object.

For this problem:

$s_{0}=0 ft$ , $v_{0}=100ft/s$ , $a=-32ft/s^{2}$

$s(t)=100t+\frac{1}{2}(-32)t^{2}$

$s(t)=100t-16t^{2}$

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Example Question #941 : Calculus

A ball is thrown straight upward from ground level. It has an initial velocity of 20m/s . At what time will the ball reach its maximum height?

Use to approximate acceleration due to gravity.

Possible Answers:

t=1

t=5

t=2

t=0

t=4

Correct answer:

t=2

Explanation:

The ball's acceleration will simply be its acceleration due to gravity, approximated with .

a(t)=-10

Integrating the acceleration function will leave you with the function for velocity.

v(t) = -10t + C

You know that has an initial velocity of 20m/s ., so v(0)=20 . Use this to find the value of the integration constant, .

20 = -10(0) + C

20=C

Therefore, our velocity funciton is v(t) = -10t +20

Finding the zero of the velocity function will give local extrema of the position function.

0= -10t +20

-20=-10t

2=t

Since the velocity of the ball changes from positive to negative about t=2 , this is a local maximum. Therefore, the ball will reach its maximum height when t=2 .

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Example Question #81 : How To Find Position

A vehicle at a position of 5 m . It accelerates from rest according to the acceleration function, a(t)

a(t)= 1.5t+4

where is time (in seconds).

Find the function representing the vehicle's position.

Possible Answers:

x(t)=9t^3+24t^2+5

x(t)=9t^3+24t^2

x(t)=0.25t^3+2t^2+5

x(t)=0.25t^3+2t^2

x(t)=9t^2+24t+5

Correct answer:

x(t)=0.25t^3+2t^2+5

Explanation:

Integrating the acceleration function will result in the function representing the vehicle's velocity. To integrate this function use the rule

$x^n\rightarrow \frac{x^{n+1}}{n+1}$ .

Therefore,

a(t)= 1.5t+4

v(t) = 0.75t^2 + 4t + C .

You know that the vehicle begins at rest, so v(0)=0 . Use this to find the value of the integration constant, .

0 = 0.75(0)^2 + 4(0) + C

0=C

Therefore, our velocity funciton is v(t) = 0.75t^2 + 4t

Integrating the velocity function will result in the function representing the vehicle's position.

v(t) = 0.75t^2 + 4t

x(t)=0.25t^3+2x^2+C'

You know that the vehicle begins at a position of 5 m , so x(0)=5 .

Use this to find the value of the integration constant.

5=0.25(0^3)+2(0^2)+C'

5=C'

Therefore,

x(t)=0.25t^3+2t^2+5

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Example Question #84 : How To Find Position

An ant is moving with a velocity given by the following function:

$v(t)=\sec^2(t)+2t+1$

What is the position function of the ant?

Possible Answers:

$\tan(t)+t^2+t+C$

$\tan(t)+\frac{t^2}{2}+t+C$

$\tan(t)+t^2+t$

$\tan(t)+t^2+1+C$

Correct answer:

$\tan(t)+t^2+t+C$

Explanation:

To find the position function of the ant, we must integrate the velocity function:

$s(t)=\int v(t)=\int (\sec^2(t)+2t+1)dt=\tan(t)+t^2+t+C$

The integration was performed using the following rules:

$\int \sec^2(x)dx=\tan(x)+C$ ,

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

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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 #944 : 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:

160

145

150

135

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 #941 : Spatial Calculus

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:

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

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

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

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

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

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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Calculus 1 : Spatial Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #941 : Spatial Calculus

Example Question #941 : Calculus

Example Question #81 : How To Find Position

Example Question #84 : How To Find Position

Example Question #941 : Spatial Calculus

Example Question #942 : Spatial Calculus

Example Question #943 : Spatial Calculus

Example Question #944 : Spatial Calculus

Example Question #945 : Spatial Calculus

Example Question #941 : Spatial Calculus

All Calculus 1 Resources

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