I need to find the velocity vector two results help please

AI Thread Summary
The discussion focuses on finding the velocity vector at t=2s for the given position function r(t). The user initially calculated v(2) as (6.8, -15.7) but was confused by another user's result of 25m/s*[0.272449, -0.62826]. It was clarified that both results are equivalent, as the second result is a simplified version of the first, factoring out 25. The user was reassured that their approach was correct when using radians for angles. The conversation concluded positively, with the user expressing happiness over the confirmation of their solution.
mimi.janson
Messages
80
Reaction score
0

Homework Statement



find the velocity vector for the time t=2s

i have given r(t)=
(50*sin((1/2)*t)*cos ((1/4)t),
-50*sin((1/2)*t)*sin((1/4)t))

Homework Equations



i also know that r(t)=
(50*sin(k*t)*cos (1/2k*t),
-50*sin(k*t)*sin(1/2k*t))
where k = 0,5s-1

The Attempt at a Solution


so i tried solving it and my attempt is shown in the attachment where i got the result
v(2)=6,8;-15,7

but someone said it has to be solved with the following results
25m/s*[0,272449, -0,62826] so I am quite confused now..

so i don't know what to do and HOPE VERY MUCH someone will check my result v(2)=6,8;-15,7 and tell me if its right or not and if so what i can do to get the right result
 

Attachments

  • x of t.png
    x of t.png
    5.1 KB · Views: 488
Physics news on Phys.org
mimi.janson said:
so i tried solving it and my attempt is shown in the attachment where i got the result
v(2)=6,8;-15,7

but someone said it has to be solved with the following results
25m/s*[0,272449, -0,62826] so I am quite confused now..

so i don't know what to do and HOPE VERY MUCH someone will check my result v(2)=6,8;-15,7 and tell me if its right or not and if so what i can do to get the right result

Actually, your answer is correct if you refer angles with radians only. There seems to be no incorrect approach here.

What someone just did is to factor out 25 for each vector. Then, simplify the values. Yours and someone's answers are equivalent.

Nicely done!
 
NasuSama said:
Actually, your answer is correct if you refer angles with radians only. There seems to be no incorrect approach here.

What someone just did is to factor out 25 for each vector. Then, simplify the values. Yours and someone's answers are equivalent.

Nicely done!

Oh GREAT ! this answer is the one that has made me most happy ever <3

:)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top