Calculus Problem: acceleration, speed, and displacement of a particle

Click For Summary
The discussion revolves around solving a calculus problem involving a particle's acceleration, speed, and displacement. The acceleration is defined as a function of time, a=A√t, with initial conditions for velocity and position. Participants emphasize the importance of using definite integrals to find speed and displacement, suggesting that the user should integrate the acceleration function while considering initial values. The integration process is clarified, highlighting the need to apply limits correctly for both velocity and position. The conversation aims to guide the user in correctly applying calculus to derive the required functions and values.
Stephanievet54
Messages
1
Reaction score
0

Homework Statement


The acceleration of a particle given a=A√t where A=2.0 m/s5/2. At t=0, v=7.5 m/s and x=0. (a) What is the speed as a function of time? (b) What is the displacement as a function of time? (c) What are the acceleration, speed, and displacement at t=5.0s.

Homework Equations

The Attempt at a Solution


I tried to integrate a=A√t, but I don't know where to plug in v and x. Any help will be greatly appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
Hi Stephanievet54 and welcome to PF.

Can you show in detail what you tried to do? Remember that in physics you deal with definite integrals. For example, if ##a=\frac{dv}{dt}##, then $$\int_{v_0}^v dv=\int_{t_0}^{t}a(t') dt'$$Note that at the lower limit of time ##t_0## the velocity is ##v_0## and at the upper limit of arbitrary time ##t## the velocity is ##v##. The integration on the right is over dummy variable ##t'##. Similar considerations apply for ##x(t)##.
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 29 ·
Replies
29
Views
2K
Replies
1
Views
1K
Replies
6
Views
3K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
953
  • · Replies 7 ·
Replies
7
Views
933
Replies
11
Views
1K