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

1. Mar 30, 2008

### HelpMeWIN123

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

1. The problem statement, all variables and given/known data

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.

2. Relevant 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?

2. Mar 30, 2008

### HelpMeWIN123

3. Mar 30, 2008

### tiny-tim

Hi HelpMeWIN123!

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

You must integrate …

4. Mar 30, 2008

### HelpMeWIN123

can you be more specific, integrate what, and integrate why?

5. Mar 30, 2008

### HelpMeWIN123

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

Last edited: Mar 30, 2008
6. Mar 30, 2008

### tiny-tim

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 … ?

7. Mar 30, 2008

### tiny-tim

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 …

(You don't like integrating, do you?

I think it would be best if you always wrote out the formulas, like v = dy/dt, for yourself at the start.)

8. Mar 30, 2008

### HelpMeWIN123

so i just integrated that further and got the answer., integrated the velocity rather.

9. Mar 30, 2008

### HelpMeWIN123

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. Mar 30, 2008

### tiny-tim

Hi HelpMeWIN123!

If in doubt … integrate anyway … even if it's a constant, you'll still get the correct result!
Sorry … what kinematics equation?

11. Mar 30, 2008

### HelpMeWIN123

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

Last edited: Mar 30, 2008
12. Mar 30, 2008

### tiny-tim

This kinematics equation only works for constant a, anyway.
You can only "shortcut" if yi (the initial distance) is zero.

Do you understand the equation ∫(dy/dt)dt = yf - yi ?

13. Mar 30, 2008

### HelpMeWIN123

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. Mar 30, 2008

### tiny-tim

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.

15. Mar 30, 2008

### HelpMeWIN123

ahhh thank you sOOOO MUCH!!!!!