Solving Curvilinear Motion: Find Velocity at t=3s

[NotaMathPerson]

The acceleration function of an object doing curvilinear motion is a = {(-0.2t)i+2j+1.5k} m/s^2, where t is in s. If its initial velocity v0 = 8i m/s, and initial position is at the origin, determine the magnitude of its velocity when t=3 s.

Badly need your help guys. Thanks!

All I see is

ax = -0.2t
ay = 4
az = 1.5

Integrating all components i have

Vx= t^2/10 m/s
Vy = 2t m/s
Vz = 1.5t m/s

@ t = 3s

I get V = 7.554 m/s

But I ignored V0 which i don't know the use of.

 

[I like Serena]

Re: Curvelinear motion

The acceleration function of an object doing curvilinear motion is a = {(-0.2t)i+2j+1.5k} m/s^2, where t is in s. If its initial velocity v0 = 8i m/s, and initial position is at the origin, determine the magnitude of its velocity when t=3 s.

Badly need your help guys. Thanks!

All I see is

ax = -0.2t
ay = 4
az = 1.5

Integrating all components i have

Vx= t^2/10 m/s
Vy = 2t m/s
Vz = 1.5t m/s

@ t = 3s

I get V = 7.554 m/s

But I ignored V0 which i don't know the use of.

Hi NotaMathPerson! :)

After integration, we should have an integration constant.
The velocity should be:
$$v_x= (t^2/10 + C_1) \text{ m/s} \\
v_y = (2t + C_2) \text{ m/s} \\
v_z = (1.5t + C_3)\text{ m/s} $$

We can substitute $t=0$ to find the initial velocity and set it equal to the given initial velocity.

 

[NotaMathPerson]

Re: Curvelinear motion

Hi NotaMathPerson! :)

After integration, we should have an integration constant.
The velocity should be:
$$v_x= (t^2/10 + C_1) \text{ m/s} \\
v_y = (2t + C_2) \text{ m/s} \\
v_z = (1.5t + C_3)\text{ m/s} $$

We can substitute $t=0$ to find the initial velocity and set it equal to the given initial velocity.

How is that going lead me to the velocity @ t=3s?

 

[I like Serena]

Re: Curvelinear motion

How is that going lead me to the velocity @ t=3s?

It allows you to find $C_1$, $C_2$, and $C_3$ (which is effectively just the initial velocity).
After that you can substitute $t=3$ and find the velocity at that time.

 

[NotaMathPerson]

Re: Curvelinear motion

It allows you to find $C_1$, $C_2$, and $C_3$ (which is effectively just the initial velocity).
After that you can substitute $t=3$ and find the velocity at that time.

You said that i can substitute t =0 to the components equation

C1 = 8, C2 = 0 C3 = 0

(t^2/10+8)i +2tj+1.5k = V

@ t = 3s
89/10i+6j+4.5k = V

Vmag = 11.639 m/s is this correct?What if The problem also ask for the magnitude of the position what should I do?

 

[I like Serena]

Re: Curvelinear motion

You said that i can substitute t =0 to the components equation

C1 = 8, C2 = 0 C3 = 0

(t^2/10+8)i +2tj+1.5k = V

@ t = 3s
89/10i+6j+4.5k = V

Vmag = 11.639 m/s is this correct?What if The problem also ask for the magnitude of the position what should I do?

It looks correct (without checking the calculations), although you left out a $t$ in the k-component of the formula for the velocity.

To find the position, integrate the equations you have for the velocity (including the missing $t$), and use the fact that the initial position is at the origin - it's the same thing!

 

[NotaMathPerson]

Re: Curvelinear motion

It looks correct (without checking the calculations), although you left out a $t$ in the k-component of the formula for the velocity.

To find the position, integrate the equations you have for the velocity (including the missing $t$), and use the fact that the initial position is at the origin - it's the same thing!

Do I have to also put an integration constant after integrating?

 

[I like Serena]

Re: Curvelinear motion

Do I have to also put an integration constant after integrating?

Yes.
In this particular case all three constants will turn out to be zero since it is given that the initial position is at the origin.

 

[HallsofIvy]

Re: Curvelinear motion

Do I have to also put an integration constant after integrating?

Didn't you learn that in Calculus I? If F is an anti-derivative of f, that is, if F'= f, then so is F+ C for any constant C.

 

[NotaMathPerson]

Re: Curvelinear motion

Yes.
In this particular case all three constants will turn out to be zero since it is given that the initial position is at the origin.

Hello.

Do you mean the sum of the three constants is zero?

Because when I integrate and let t =0. I get Ca + Cb + Cc = 0 . What do you mean by all three constants are zero? Do you mean individually?

 

[HallsofIvy]

No! We did not sat that and did not mean that! The three components of the vector are separate functions. They are not to be added.

 

[Gustavo1]

Re: Curvelinear motion

Hello.

Do you mean the sum of the three constants is zero?

Because when I integrate and let t =0. I get Ca + Cb + Cc = 0 . What do you mean by all three constants are zero? Do you mean individually?

Hi, Sr, I would like to know if you finally resolve this problem, I really do not know how to finish...do you the solution?
I am interested in this problem becouse I saw it in this link:

https://www.youtube.com/watch?v=gQMijT0hZdk&index=6&list=PLLbvVfERDon1xk3wGaYfXSmGa1u83mGn-

in the minute 6:37

I really appreciate.