How Does Acceleration Affect Velocity in 1D Motion?

[rudransh verma]

Homework Statement

The acceleration time graph of a particle moving in straight line is given below. The velocity of particle at t=0 is 2 m/s. The velocity after 2s will be?

Relevant Equations

Area under the curve will give velocity.

$v=\frac12 *1*4+\frac12*1*4= 4 m/s$ but the answer is wrong.

 

[BvU]

That's because your relevant equation is wrong. What is the velocity at t=6 if the area is zero ?

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[rudransh verma]

That's because your relevant equation is wrong. What is the velocity at t=6 if the area is zero ?

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If area is zero means no acceleration. That means velocity v=u, the initial velocity.

 

[BvU]

So ...

 

[rudransh verma]

So ...

So v=2m/s. But here there is acceleration and equal acceleration and de acceleration. So I think it should v=u at t=2sec

 

[mjc123]

There is no deceleration. The magnitude of the acceleration increases to a maximum then decreases, but it is always positive.

The statement in your first post is wrong. Area under the curve equals change in velocity. If the initial velocity is 2 m/s, then the final velocity is...?

 

[BvU]

There is no deceleration. The magnitude of the acceleration increases to a maximum then decreases, but it is always positive.

The statement in your first post is wrong. Area under the curve equals change in velocity. If the initial velocity is 2 m/s, then the final velocity is...?

Giving it all away, eh ?
Much better if @rudransh verma discovers this for him/her self !

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[mjc123]

Giving it all away, eh ?
Much better if @rudransh verma discovers this for him/her self !

It's stated in the image posted in post #1. I'm just making it more explicit in words. I assumed that's legitimate as OP has made an attempt at the question and explained their thinking, it's OK to point out where they're mistaken. I'm not working out the answer for them.

 

[rudransh verma]

There is no deceleration. The magnitude of the acceleration increases to a maximum then decreases, but it is always positive.

Can you give me an example where there is equal acceleration and de acceleration via graph?

Area under the curve equals change in velocity.

But I have read area under curve gives velocity. Is it because initial velocity is not equal to zero.

 

[Lnewqban]

Can you give me an example where there is equal acceleration and de acceleration via graph?

But I have read area under curve gives velocity. Is it because initial velocity is not equal to zero.

Could you produce a graph showing how the velocity changes respect to time for this specific problem?
It is difficult not to get confused by graphs of acceleration versus time, because in practical life we are used to constant values of acceleration (horizontal lines in a a-t graph), like when an object falls (at gravity acceleration) or when a car decreases its velocity at a steady rate (slowly braking before a traffic light).

 

[rudransh verma]

Could you produce a graph showing how the velocity changes respect to time for this specific problem?

 

[BvU]

The straight line graph you show would mean the acceleration $dv\over dt$ is constant ! Post #1 clearly doesn't have that.

equal acceleration and de acceleration

You are mistaken here: from time 0 to 1 there is acceleration and from time 1 to 2 there is acceleration too !
As you calculate correctly, the area under the $a(t)$ curve is +4 m/s.

Hint: Check out post #6 once more

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[Lnewqban]

 

[BvU]

Giving it away, eh ?

 

[rudransh verma]

View attachment 295898

HOw did you do that with0ut a function ?

 

[Lnewqban]

We don't need a function to draw an approximate shape.
What is important to me is that the graph helps you understand what is approximately happening to the values of the velocity as time goes by.
Re-visit post #12 above: dv/dt is the slope of ...?

 

[jbriggs444]

Can you give me an example where there is equal acceleration and de acceleration via graph?

Here is a period of acceleration followed by an equal period of deceleration depicted on an acceleration versus time graph.

You can see that the area "under" this curve is zero.

 

[rudransh verma]

Here is a period of acceleration followed by an equal period of deceleration depicted on an acceleration versus time graph.
View attachment 295941
You can see that the area "under" this curve is zero.

 

[jbriggs444]

I see that you have produced a graph. What are you graphing (what versus what) and what is it intended to depict?

Possibly it is acceleration versus time and you have answered your own request by producing a graph depicting an acceleration followed by an equal period of deceleration.

 

[rudransh verma]

I see that you have produced a graph. What are you graphing (what versus what) and what is it intended to depict?

I am trying to understand that in OP graph and this graph that the acceleration is first increasing then decreasing. I am unable to understand what does it do to velocity? Is velocity increasing at a rate and then increasing at a lower rate but not getting deaccelerated?
Similarly in above graph first the acceleeration is constant then decreasing then zero and then its -ve.

 

[jbriggs444]

I am trying to understand that in OP graph and this graph that the acceleration is first increasing then decreasing.

In the graph in the OP, acceleration is first increasing and then decreasing.

In the graph you most recently posted, acceleration is first positive and constant then positive and decreasing. It becomes negative and then remains negative and constant.

In the graph that I posted, acceleration is first positive and constant. It then changes discontinuously and becomes negative and constant.

Those three situations are all different. The three graphs look different.

I am unable to understand what does it do to velocity? Is velocity increasing at a rate and then increasing at a lower rate but not getting deaccelerated?

While acceleration is positive, velocity becomes more positive.
While acceleration is negative, velocity becomes more negative.

 

[rudransh verma]

While acceleration is positive, velocity becomes more positive.
While acceleration is negative, velocity becomes more negative.

Yeah. But when acceleration becomes less positive then the rate of change of velocity in positive direction decreases. For example The velocity of a car will increase continuously but its just that it will take more time. At some point the acceleration will become zero so no increase in velocity and constant v. After that as the -a appears with + direction of displacement deacceleration starts, cars speed decreases.
Right