Solving 1-D Motion Questions with Constant Velocity and Acceleration

[Shing]

Hi guys, I am self-studying physics, but I am confused for something following:

given 1-d motion of an object o, we can use these equations to describe the motion of it.
\overline{v} is a constant, so is\overline{a} .

1)x_t=x_0+\overline{v}t

2)v_t=v_0+\overline{a}t

3)d(x_t,x_o)=v_ot+\frac{1}{2}\overline{a}t^2

if we take an rest reference frame of o

then how will those equations be changed?
here is how I think:
in case I,
there is no acceleration,
x_t=x_0+\overline{v}t=x_0 if v_0=\overline{v}
question 1. what ifv_0>\overline{v}? and v_0<\overline{v}
should it be written as: x_t=x_o+|\overline{v}-v_r|t??



in case II,
there is acceleration,
I am really confused by this case,
since the textbook told me there is no relative motion when it comes to acceleration. And I have no idea WHY.
and after reading Newton's bucket, I am even more confused.

I have no idea how I should write the two equations 2)&3)
question 2.the acceleration seems to be changed... but how?

thanks for your reading, please help me =)

 

[Doc Al]

Hi guys, I am self-studying physics, but I am confused for something following:

given 1-d motion of an object o, we can use these equations to describe the motion of it.
\overline{v} is a constant, so is\overline{a} .

1)x_t=x_0+\overline{v}t

2)v_t=v_0+\overline{a}t

3)d(x_t,x_o)=v_ot+\frac{1}{2}\overline{a}t^2

Those look like kinematic equations for constant acceleration. But what you mean by the lines over the symbols? (\overline{v} & \overline{a} ?)

If the acceleration is constant (and non-zero), how can the speed be constant also?

And if you take a frame co-moving with the particle, the equation of motion becomes: x = x0 (it doesn't move in its own frame).

Perhaps you can give us more background and rephrase your question.

 

[Shing]

my bad, sorry for my poor expression

1) --> o's rest reference frame, no acceleration.
2) and 3) --> when o has an constant acceleration.

in case I,
if we take the rest reference frame of o, and it doesn't have acceleration.
So we will observe a rest o

and how we express in equation is:
x_t=x_0+(v-v_r)t=x_o

right?

and in case II, o has an constant acceleration

and how should we express the equation if when we take the reference with the same speed as the v_o?

v_t=v_o-v_r+at=at

and the speed of the reference doesn't affect how we observe the acceleration?

and thinking further ,
if an object's motion can be expressed by \frac{d^nx}{dt^n}
would how we observe the object be affected by the taken reference frame expressed as \frac{d^{n-1}x}{dt^{n-1}}?

 

[Doc Al]

I think you're asking: Take a standard reference frame and a second reference frame moving at speed V with respect to the first.

If an object has a constant speed V in the first frame, its position will be: x = x0 + Vt in that frame. From the second frame it will be x' = x - Vt = x0 and v' = v -V = 0.

It the particle is accelerating, x = x0 + v0t + .5at^2 and v = v0 + at. In the second frame, x' = x -Vt = x0 + v0t + .5at^2 -Vt = x0 + .5at^2 and v' = v - V = v0 + at - V = at.

Which I think is just what you are saying. Right?

 

[Shing]

I think you're asking: Take a standard reference frame and a second reference frame moving at speed V with respect to the first.

If an object has a constant speed V in the first frame, its position will be: x = x0 + Vt in that frame. From the second frame it will be x' = x - Vt = x0 and v' = v -V = 0.

It the particle is accelerating, x = x0 + v0t + .5at^2 and v = v0 + at. In the second frame, x' = x -Vt = x0 + v0t + .5at^2 -Vt = x0 + .5at^2 and v' = v - V = v0 + at - V = at.

Which I think is just what you are saying. Right?

Yes, thank you for your help =)
but I also would like to know that:
if the object is accelerating, can we take rest reference frame of it?

 

[Doc Al]

but I also would like to know that:
if the object is accelerating, can we take rest reference frame of it?

I think you are asking: If an object is accelerating, can we view things from a frame in which the object is at rest?

Sure, why not?

 

[Shing]

I think you are asking: If an object is accelerating, can we view things from a frame in which the object is at rest?

Sure, why not?

but if we view things from a frame in which the object is at rest, but actually it is accelerating.
we still can feel we are accelerating, pushed back.
And will we still think that "we" and "that object" are still at rest as how we observe?

 

[arildno]

but if we view things from a frame in which the object is at rest, but actually it is accelerating.
we still can feel we are accelerating, pushed back.
And will we still think that "we" and "that object" are still at rest as how we observe?

And such frames are NON-INERTIAL reference frames, and the transition from them to an inertial reference frame is quite different than from another inertial reference frame.

The relativity of inertial reference frames means that the "laws of physics" are indistinguishable between two such (i.e, the same forces and accelerations will be calculated from either frame), whereas an accelerated reference frame will calculate another set of forces and accelerations than those calculated within an inertial frame.

 

[Doc Al]

but if we view things from a frame in which the object is at rest, but actually it is accelerating.
we still can feel we are accelerating, pushed back.
And will we still think that "we" and "that object" are still at rest as how we observe?

Just to add to what arildno already explained...

If all you care to do is describe the motion (kinematics), then there's no problem at all in using an accelerated frame. But as soon as you want to talk about the physics (dynamics), then you must face the consequences of using a non-inertial frame of reference. In order to apply Newton's laws in such a frame, one must introduce inertial pseudo-forces to account for the fact that you are using such a frame.

 

[Shing]

Thanks to DocAl & Arildno!

but I still have some questions,

if there are two reference frames for an object, let's say, Y

If Y is accelerating in the first reference frame.
And compared to the first reference frame, the second reference frame is accelerating too (a lower <br /> \frac{d^2x}{dt^2}<br /> .)
from the second reference frame, if we measure the acceleration of Y.
Will the acceleration be different to what we measure from the first?
I've draw a graph, and I found the acceleration will be different from different frames, is that correct?

 

[Doc Al]

Will the acceleration be different to what we measure from the first?
I've draw a graph, and I found the acceleration will be different from different frames, is that correct?

Sure it will be different.

As you know, any two inertial frames will measure the object as having the same acceleration (at least non-relativistically).

But if one of the frames is accelerating, it will measure a different acceleration for the object. An obvious example is if the second frame is accelerating at the same rate as the object--in that case the object would appear to have no acceleration at all.