Simple velocity vector problem; IS MY SOLUTION MANUAL WRONG OR AM I?

In summary, the particle has a velocity of 5i at t=0, and moves with a varying acceleration given by a= 6 Sqrt(t) j.
  • #1
HelpMeWIN123
20
0
Simple velocity vector problem; IS MY SOLUTION MANUAL WRONG OR AM I?!

Homework Statement



A particle starts from the origin with velocity 5i at t=0, and moves in the xy plane with a varying acceleration given by a= 6 Sqrt(t) j, where t is in s.
a.) Determine the vector velocity of the particle as a function of time.

Homework Equations


Vi=5i; a = 6 sqrt(t)j
Vf=Vi+at
Vf = Vfx +Vfy = (Vix+axt)i+(Viy+ayt)j
=(5+0)i + (0+6sqrt(t)*t)j
=5i +6t^3/2 j

The solution manual states the answer as Vf=5i +4t^3/2 j
Whos right here?
 
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  • #2
anyone? please? pretty please?
 
  • #3
HelpMeWIN123 said:
Vf = Vfx +Vfy = (Vix+axt)i+(Viy+ayt)j

Hi HelpMeWIN123! :smile:

No no no … that formula only works if a is constant! :frown:

You must integrate … :smile:
 
  • #4
can you be more specific, integrate what, and integrate why?
 
  • #5
AHHH you're right, and a genius, and a scholar.
Acceleration is not constant, so we can't use those formulas for constant acceleration.
whenever they give you an a with respect to time, this is a variability acceleration, so in order to achieve V we must integrate a.

Thanks.
But now if I'm trying to find the position with respect to time;
it would be : 5t i + [0 +0t + (1/2) (6 sqrt(t)*t^2) ]j
isn't it?
which gives me 3t^5/2 j? this isn't right either ahhhhhh since acceleration isn't constant..
so i took the velocity vector 5i +4t^3/2j
and differentiated it to achieve a, for each component giving me, a= 0i +6 sqrt(t)j which give me right back to where I was in terms of the previous plug-in. grr
 
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  • #6
ok … acceleration is rate of change of velocity.

In other words: a = dv/dt.

So if you integrate, you get ∫adt = ∫(dv/dt)dt = vf - vi.

If a is constant, that's just the usual at = vf - vi formula.

But in this case, a isn't constant.

So … ? :smile:
 
  • #7
ah … just seen your last post.

ok … same thing … speed is rate of change of distance.

In other words: v = dy/dt.

So if you integrate, you get ∫vdt = ∫(dy/dt)dt = yf - yi.

So just integrate v … :smile:

(You don't like integrating, do you? :rolleyes:

I think it would be best if you always wrote out the formulas, like v = dy/dt, for yourself at the start.)
 
  • #8
so i just integrated that further and got the answer., integrated the velocity rather.
 
  • #9
Yeah you're right, I guess here I just didn't understand how to differentiate (figuratively) the different of when I have constant a or constant v versus not, the thing is once I have the velocity that I got from integrating my acceleration, why can't i just plug that into my kinematics equation? kinematics eq. says nothing about having constant v
 
  • #10
Hi HelpMeWIN123! :smile:

If in doubt … integrate anyway … even if it's a constant, you'll still get the correct result! :smile:
HelpMeWIN123 said:
the thing is once I have the velocity that I got from integrating my acceleration, why can't i just plug that into my kinematics equation? kinematics eq. says nothing about having constant v

Sorry … what kinematics equation? :confused:
 
  • #11
the kinematics equation i used upstairs; yf = yi +viy *t +1/2 ay t^2
also how does ∫(dy/dt)dt = yf - yi
could i just shortcut and say ∫(dy/dt)dt = yf?
I think I'm utterly lost here...isn't dy/dt already vf-vi when change in time approaches 0
 
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  • #12
HelpMeWIN123 said:
the kinematics equation i used upstairs; yf = yi +viy *t +1/2 ay t^2

HelpMeWIN123 said:
kinematics eq. says nothing about having constant v

This kinematics equation only works for constant a, anyway. :frown:
also how does ∫(dy/dt)dt = yf - yi
could i just shortcut and say ∫(dy/dt)dt = yf?
I think I'm utterly lost here.

You can only "shortcut" if yi (the initial distance) is zero.

Do you understand the equation ∫(dy/dt)dt = yf - yi ?
 
  • #13
well i was tryign to make sense of it in the fact that
V= dy/dt
Vdt = dy, and dy = yf-yi or change in y
then ∫V dt = yf - yi is where i get stuck because how you can just integrate one side and how do you have the wherewithall to do this.
 
  • #14
No … it's because y = ∫(dy/dt)dt.

(Perhaps that looks more familiar as ∫y´= y ?)

So ∫(dy/dt)dt between a and b

= [y] between a and b,

= y(a) - y(b) …
which in dynamics we usually write yf - yi. :smile:
 
  • #15
ahhh thank you sOOOO MUCH!
 

1. Is my solution manual wrong?

It is possible that your solution manual contains errors. However, it is always a good idea to double check your work and make sure that you have followed all the necessary steps to arrive at the correct solution. If you are still unsure, it is best to consult with your instructor or a fellow classmate for clarification.

2. Why does my solution differ from the one in the manual?

There are many factors that can contribute to a different solution, such as rounding errors, incorrect assumptions, or mistakes in calculations. Make sure to carefully review your work and check for any discrepancies. If you are still unsure, seek guidance from your instructor.

3. How do I know if my solution is correct?

One way to check the accuracy of your solution is to plug your values back into the original problem and see if it satisfies all the given conditions. You can also compare your solution to the one in the manual or ask your instructor for feedback.

4. Am I missing any steps in my solution?

It is possible that you may have overlooked a step or made a mistake in your calculations. Double check your work and make sure you have followed all the necessary steps. You can also ask a classmate or your instructor to review your solution and provide feedback.

5. What should I do if I am still confused about the problem?

If you are still unsure about the problem or your solution, do not hesitate to seek help from your instructor or a tutor. They can provide further explanation and guide you towards a better understanding of the problem. It is important to ask for help when needed in order to fully grasp the concept and improve your problem-solving skills.

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