Kinematics motion-setting up instantaneous velocity function

[negation]

Homework Statement

An object starts moving in a straight line from position xi, at time t =0, with velocity vi. Its acceleration is given by a = ai + bt, where ai and b are constant. Find expressions for
a)instantaneous velocity
b) position as functions of time.

Homework Equations

none

The Attempt at a Solution

 

[CFede]

The equation you have there (xf=xi+vi t+0.5 ai t^2) is only true for constant acceleration. You should workout the instantaneous velocity by integrating the acceleration with respect to time. THen do the same thing to get the position.

 

[negation]

The equation you have there (xf=xi+vi t+0.5 ai t^2) is only true for constant acceleration. You should workout the instantaneous velocity by integrating the acceleration with respect to time. THen do the same thing to get the position.

But the acceleration that is given, a = ai + bt, is already a function of time.

 

[Chestermiller]

But the acceleration that is given, a = ai + bt, is already a function of time.

How is the derivative of the velocity with respect to time related to the acceleration?

 

[negation]

How is the derivative of the velocity with respect to time related to the acceleration?

I'm lost. Could someone provide an exposition?

 

[negation]

How is the derivative of the velocity with respect to time related to the acceleration?

Differentiating displacement with respect to time gives me instantaneous velocity, no?

 

[Chestermiller]

I'm lost. Could someone provide an exposition?

$$\frac{dv}{dt}=a=a_i+bt$$
Do you know how to integrate this differential equation?

 

[negation]

$$\frac{dv}{dt}=a=a_i+bt$$
Do you know how to integrate this differential equation?

I do.

but why can't I set up a displacement equation as a function of time as like in my first post and differentiate it?
What's the rational?

 

[Chestermiller]

I do.
but why can't I set up a displacement equation as a function of time as like in my first post and differentiate it?
What's the rational?

The rationale is that the equation in your first post is incorrect if the acceleration is a function of time. So it's of no use in this problem. You need to start with the acceleration and work backwards by getting the velocity first and then the displacement. Compare what you get doing it this way your the equation in your first post.

 

[negation]

The rationale is that the equation in your first post is incorrect if the acceleration is a function of time. So it's of no use in this problem. You need to start with the acceleration and work backwards by getting the velocity first and then the displacement. Compare what you get doing it this way your the equation in your first post.

I have inserted the antiderivative in the above post after you quote.

 

[Chestermiller]

I have inserted the antiderivative in the above post after you quote.

Yes. That's the increase in velocity. If you want the velocity itself, you need to add in the initial velocity. Then, integrate the resulting equation to get the location x.

 

[negation]

yes. That's the increase in velocity. If you want the velocity itself, you need to add in the initial velocity. Then, integrate the resulting equation to get the location x.

 

[Chestermiller]

View attachment 64973

This was done correctly, but the problem statement asked for the position as a function of time.

 

[negation]

This was done correctly, but the problem statement asked for the position as a function of time.

Alright. This is delta x. To get xf, xf= xi+ delta x.

 

[Chestermiller]

Alright. This is delta x. To get xf, xf= xi+ delta x.

Yeah! Excellent!