The velocity of the particle as a function of time

Click For Summary
The discussion focuses on solving for the velocity of a particle given its acceleration function a(t) = bt y + c e^(kt) z. A user expresses confusion about the correct form of the velocity equation, questioning whether it should be (c/k) * exp(kt) without the exponent notation. They seek clarification on their approach and whether their interpretation of the problem is incorrect. The conversation highlights the need for understanding the integration of acceleration to find velocity and the importance of correctly applying mathematical notation. The thread emphasizes the complexities involved in solving motion equations in physics.
yesmale4
Messages
41
Reaction score
1
Homework Statement
The acceleration of the particle is given by a(t)=bt y + c e kt z.
It is known that at t=0 the velocity of the particle is by v(0)= d x + c/k z and its position is r(0)= c/k2 z.
b, c, d and k are constants.
Relevant Equations
v\left(t\right)=\int \:a\left(t\right)dt
sa.png

this is how i try to solve it:
mm.jpeg


can someone please help me with that because i don't know what I am doing worng here.
 

Attachments

  • sa.png
    sa.png
    11.4 KB · Views: 136
Physics news on Phys.org
Isn't the problem that it should be
Code:
(c/k)*exp(k*t)
(without the ^)?
 
DrClaude said:
Isn't the problem that it should be
Code:
(c/k)*exp(k*t)
(without the ^)?
Thank you very much !
 
yesmale4 said:
Homework Statement:: The acceleration of the particle is given by a(t)=bt y + c e kt z.
It is known that at t=0 the velocity of the particle is by v(0)= d x + c/k z and its position is r(0)= c/k2 z.
b, c, d and k are constants.
Relevant Equations:: v\left(t\right)=\int \:a\left(t\right)dt

View attachment 297864
this is how i try to solve it:
View attachment 297866

can someone please help me with that because i don't know what I am doing wrong here.
Is it telling you it is wrong or is that just what you think?
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
757
  • · Replies 13 ·
Replies
13
Views
662
Replies
17
Views
1K
Replies
8
Views
1K
Replies
23
Views
2K
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K