Dynamic Acceleration Problem: Finding Velocity at t = 6.5 s

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Homework Help Overview

The discussion revolves around a dynamic acceleration problem where participants are trying to find the velocity of a particle given its acceleration function a = 5.0t and an initial velocity at t = 2.0 s. There is also a related problem concerning the position of a point given its acceleration and initial conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between acceleration and velocity, considering both average and instantaneous acceleration. There are attempts to apply calculus, particularly integration, to find velocity from acceleration. Questions arise about the role of constants of integration and how to apply them correctly.

Discussion Status

Some participants have provided guidance on using integration to find velocity and position from acceleration. There is an ongoing exploration of how to handle constants of integration and the implications of initial conditions. Multiple interpretations of the problems are being discussed without a clear consensus on the final answers.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is also a focus on understanding the application of derivatives and integrals in the context of dynamic systems.

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The acceleration of a particle along an x-axis is a = 5.0t, with t in seconds and a in meters per second squared. At t = 2.0 s, its velocity is +13 m/s. What is its velocity at t = 6.5 s?




I don't really know how to go about this, as it is a dynamic acceleration, any help would be greatly appreciated

so far all I have got is a=deltaV/deltaT
 
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B-80 said:
so far all I have got is a=deltaV/deltaT
That's OK for average acceleration, but for instantaneous acceleration use a derivative: [itex]a = dv/dt[/itex].

Given acceleration, how do you find velocity? Hint: Use a little calculus.
 
well I would multiply by the time passage, but again this only works for constant acell right? my problem is the derivative, I know how to do them, but i don't understand how that applies. the derivative of a is 5 right? that's great, but what do I do after that
 
If you were given v, you'd take its derivative to find a. But you have the opposite situation: You are given a and need to find v. What do you do?
 
integrate? so 2.5t^2=v
 
B-80 said:
integrate? so 2.5t^2=v
Good! But don't forget the constant of integration.
 
wow awesome thanks a lot.
 
Finding position

I have run across the same type of problem, but I am being asked what the position is. Below is what was given.

Acceleration of a point is a=20t m/s^2. When t=0, s=40 meters and v=-10 m/s. What is the position at t=3 sec?


I've used a=dv/dt to get the velocity, v=80 m/s.

I'm not sure were to start to get the position. Can you advise?

Thank you.
 
Using a=dv/dt, you integrated to get the velocity. But v = dx/dt, so integrate once again to get the position.
 
  • #10
For velocity I got v=10t^2 + c, and pluged 3 in for t and -10 in for c. This gave me v=80 m/s.

If i integrate once more to get position I get s=(10t^3)/3+ct.

Am I correct with the constant?
Plug 3 in for t, and -10 in for c again.

I think I am getting confused with the constant.
 
  • #11
Bingo1915 said:
For velocity I got v=10t^2 + c, and pluged 3 in for t and -10 in for c. This gave me v=80 m/s.
That's the velocity at t = 3; in general it's: v=10t^2 - 10.

If i integrate once more to get position I get s=(10t^3)/3+ct.
But you forgot the new constant of integration. (Hint: You are given the position at t = 0.)
 
  • #12
Thats what I was forgetting.

Therefore s=(10t^3)/3+ct+d.

Where
t=3
c=-10
d=40

Thanks for the help.
 

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