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Calculus 1 : How to find position

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Example Questions

Example Question #935 : Calculus

Given that the final acceleration is $32 \frac{m}{s^2}$ , and that the final position is 44m with a velocity of $45\frac{m}{s}$ after 10 seconds, find the initial position at t=0 .

Possible Answers:

1194m

-1564m

1040m

Correct answer:

1194m

Explanation:

We have to start from acceleration and integrate back to position to determine the initial position at t=0 .

$V(t)=\int 32= 32t+V_0$

Since we know that the velocity at t=10 is $45\frac{m}{s}$ , we plug it in to determine the initial velocity

V(10)=45=32(10)+V_0 , V_0=-275

V(t)=32t-275

Now that we know the function for velocity, we can integrate it to find position

$P(t)= \int V(t)=16t^2-275t+P_0$

Since we also know that at t=10 that the position is at 44m , we use it to find initial position at t=0

P(10)=44=16(10)^2-275(10)+P_0 , P_0= 1194

Therefore, at t=0, the initial position is 1194m.

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Example Question #81 : How To Find Position

Find the position equation of the particle, given the following information:

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

s(0)=0

Possible Answers:

$\cos(t)+t^2+1$

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

$-\sin(t)+t^2$

$\sin(t)+t^2$

Correct answer:

$\sin(t)+t^2$

Explanation:

In order to find the position equation, we must integrate the velocity function and then plug in the initial condition to solve for C:

$\int (\cos(t)+2t)dt=\sin(t)+t^2+C$

We used the following rules for integration:

$\int\cos(x)dx=\sin(x)+C$ , $\int x^ndx=\frac{1}{n+1}x^{n+1}+C$

Now, plug in t=0 into the position equation and solve for C:

s(0)=0=0+0+C

Thus, C=0, and our final answer is

$\sin(t)+t^2$

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Example Question #933 : Spatial Calculus

The velocity of a particle is given by the function v(t)=2e^t-3sin(t) .

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

Possible Answers:

$2e^{\pi}-5$

$2e^{\pi}-3$

$2e^{\pi}$

$2e^{\pi}+2$

$2e^{\pi}-8$

Correct answer:

$2e^{\pi}-8$

Explanation:

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

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

Velocity is given as:

v(t)=2e^t-3sin(t)

So the position function is:

x(t)=2e^t+3cos(t)+C

The integration of constant is found by using the initial condition provided:

x(0)=0=2e^0+3cos(0)+C

C=-5

Therefore

x(t)=2e^t+3cos(t)-5

$x(\pi)=2e^{\pi}-3-5$

$x(\pi)=2e^{\pi}-8$

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Example Question #81 : Position

The velocity of a particle is given by the function $v(t)=6t+e^{-3t}$ . If the particle has an initial position of , what will its position be at time t=2 ?

Possible Answers:

$\frac{35+e^{-6}}{3}$

$\frac{37-e^{-6}}{3}$

$12+\frac{e^{-6}}{3}$

$12-\frac{e^{-6}}{3}$

Correct answer:

$\frac{37-e^{-6}}{3}$

Explanation:

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

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

For the velocity function

$v(t)=6t+e^{-3t}$

The position function is:

$x(t)=3t^2-\frac{e^{-3t}}{3}+C$

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

$x(0)=3(0)^2-\frac{e^0}{3}+C$

$C=\frac{1}{3}$

So the definite integral is:

$x(t)=3t^2-\frac{e^{-3t}}{3}+\frac{1}{3}$

$x(2)=12-\frac{e^{-6}}{3}+\frac{1}{3}$

$x(2)=\frac{37-e^{-6}}{3}$

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Example Question #82 : How To Find Position

Find the position function of a ball thrown by a person tall if it's initial velocity is $v_{0}=40 m/s$ and its acceleration is $a=-9.8m/s^{2}$ .

Possible Answers:

$s(t)=2+40t-4.9t^{2}$

s(t)=2

s(t)=2+40t

$s(t)=2+40t-9.8t^{2}$

$s(t)=40t-9.8t^{2}$

Correct answer:

$s(t)=2+40t-4.9t^{2}$

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:

$s_{0}=2 m$ , $v_{0}=40m/s$ , $a=-9.8m/s^{2}$

$s(t)=2+40t+\frac{1}{2}(-9.8)t^{2}$

$s(t)=2+40t-4.9t^{2}$

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Example Question #939 : Calculus

Find the position of a ball after seconds if it is thrown by a person tall and its initial velocity is $v_{0}=40 m/s$ and its acceleration is $a=-9.8m/s^{2}$ .

Possible Answers:

s=79.5

s=40

s=5

s=2

s=-9.8

Correct answer:

s=79.5

Explanation:

To find the position of the object after a certain time, we first must find the position function of the object, then solve the position function at that given time.

For this problem:

$s_{0}=2 m$ , $v_{0}=40m/s$ , $a=-9.8m/s^{2}$ , t=5s

$s(t)=2+40t+\frac{1}{2}(-9.8)t^{2}$

$s(t)=2+40t-4.9t^{2}$

$s(5)=2+40(5)-4.9(5)^{2}=2+200-122.5=79.5$

After seconds, the ball has travelled 79.5 meters.

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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:

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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Calculus 1 : How to find position

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #935 : Calculus

Example Question #81 : How To Find Position

Example Question #933 : Spatial Calculus

Example Question #81 : Position

Example Question #82 : How To Find Position

Example Question #939 : Calculus

Example Question #941 : Spatial Calculus

Example Question #941 : Calculus

Example Question #81 : How To Find Position

Example Question #84 : How To Find Position

All Calculus 1 Resources

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