The position function from given velocity or acceleration function

In summary: Similarly, for r(0) = <0, 4, -2>, you can substitute 0 for t in the position function and solve for the constants in the same way. The key is to use the given information for v(0) and r(0) to determine the values of the constants in the velocity and position functions.
  • #1
2x2lcallingcq
9
0
Problem:

Find the position function from the given velocity or acceleration function.

a(t) = <e^-3t,t,sint>, v(0)=<4,-2,4>, r(0)=<0,4,-2>

Solution:

To find the answer the integral must be taken...

Integral of a(t) = <-(1/3)e^(-3t), (1/2)t^2, -cos(t)>

Since taking it with respect to t, it becomes velocity


(acceleration = x/t/t distance over time 2)

Integral of acceleration with respect to t is... (xt^-1, x/t)

Since that is just the velocity function, you do have an initial velocity at v(0)...(time at 0)
adding these to each part of the integral.

The velocity function is this: <-(1/3)e^(-3t)+4, (1/2)t^2-2, -cos(t)+4>

Again, the integral must be taken and it becomes the position function

x(t) = <(1/9)e^(-3t)+4t, (1/6)t^3-2t, 4t-sin(t)>

And add constants again r(0)

therefore the answer r(t) (x(t)) = <(1/9)e^(-3t)+4t+0, (1/6)t^3-2t+4, 4t-sin(t)-2>


BUT my teacher said the first and the third portions are wrong? I do not understand how.
 
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  • #2
2x2lcallingcq said:
Problem:

Find the position function from the given velocity or acceleration function.

a(t) = <e^-3t,t,sint>, v(0)=<4,-2,4>, r(0)=<0,4,-2>

Solution:

To find the answer the integral must be taken...

Integral of a(t) = <-(1/3)e^(-3t), (1/2)t^2, -cos(t)>
Put the constants in right away.
v(t) = <-(1/3)e-3t + c1, (1/2)t2 + c2, -cos(t) + c3>
You're given that v(0) = <4, -2, 4>, so you can solve for the three constants. It's not a simple matter of adding them in as you did.

Do the same thing when you find r(t).
2x2lcallingcq said:
Since taking it with respect to t, it becomes velocity


(acceleration = x/t/t distance over time 2)

Integral of acceleration with respect to t is... (xt^-1, x/t)

Since that is just the velocity function, you do have an initial velocity at v(0)...(time at 0)
adding these to each part of the integral.

The velocity function is this: <-(1/3)e^(-3t)+4, (1/2)t^2-2, -cos(t)+4>

Again, the integral must be taken and it becomes the position function

x(t) = <(1/9)e^(-3t)+4t, (1/6)t^3-2t, 4t-sin(t)>

And add constants again r(0)

therefore the answer r(t) (x(t)) = <(1/9)e^(-3t)+4t+0, (1/6)t^3-2t+4, 4t-sin(t)-2>


BUT my teacher said the first and the third portions are wrong? I do not understand how.
 
  • #3
Mark44 said:
Put the constants in right away.
v(t) = <-(1/3)e-3t + c1, (1/2)t2 + c2, -cos(t) + c3>
You're given that v(0) = <4, -2, 4>, so you can solve for the three constants. It's not a simple matter of adding them in as you did.

Do the same thing when you find r(t).

so I just put the constants in and then solve? I'm still confused.
 
  • #4
You are given that v(0) = <4, -2, 4>, and from your work you can substitute 0 for t to get v(0) = <-(1/3)e-3*0 + c1, (1/2)02 + c2, -cos(0) + c3>.

From this you can solve for the constants.
 

1. What is a position function?

A position function is a mathematical function that describes the position of an object at a given time. It is typically represented as a function of time, and can be used to calculate the location of an object at any point in time.

2. How is a position function related to velocity and acceleration?

A position function is related to velocity and acceleration through calculus. The position function is the integral of the velocity function, and the velocity function is the derivative of the position function. Similarly, the velocity function is the integral of the acceleration function, and the acceleration function is the derivative of the velocity function.

3. How do you find the position function from a given velocity or acceleration function?

To find the position function from a given velocity function, you can integrate the velocity function with respect to time. Similarly, to find the position function from a given acceleration function, you can integrate the acceleration function twice with respect to time. This will give you the position function in terms of time.

4. What is the difference between a position function and a displacement function?

A position function and a displacement function both describe the position of an object, but they differ in their starting point. A position function gives the position of an object at any given time, while a displacement function gives the change in position from a specific starting point.

5. Can a position function be negative?

Yes, a position function can be negative. A negative position simply indicates that the object is located in the negative direction from the reference point. This is often seen in physics problems involving motion on a number line or coordinate plane.

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