Problems with calculating the acceleration

  • Thread starter Thread starter PhyIsOhSoHard
  • Start date Start date
  • Tags Tags
    Acceleration
Click For Summary
The discussion revolves around calculating the acceleration of a block moving 11 meters in 5 seconds, leading to confusion over three different calculated values. The key point is the distinction between average velocity and instantaneous velocity, with the average velocity being 11/5 m/s. It is clarified that since the block starts from rest, the initial velocity (V0x) is zero, which affects the choice of equations used for the calculation. The only applicable formula for this scenario is the one that incorporates distance, time, and acceleration without relying on instantaneous velocity. Ultimately, understanding these concepts resolves the discrepancies in the acceleration calculations.
PhyIsOhSoHard
Messages
157
Reaction score
0

Homework Statement


A block moves 11 meters in 5 seconds on a straight line. Calculate the acceleration.

Homework Equations


v^2_x=v^2_{0x}+2a_x(x-x_0)

v_x=v_{0x}+a_xt

x=x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^2

The Attempt at a Solution


a_x=\frac{(\frac{11m}{5s})^2}{2\cdot 11m}=0.22m/s^2

a_x=\frac{(\frac{11m}{5s})}{5s}=0.44m/s^2

a_x=\frac{11m}{1/2\cdot (5s)^2}=0.88m/s^2

Why am I getting 3 different answers? In my textbook, it lists the equations as being practically the same (straight-line motion with constant acceleration).
 
Physics news on Phys.org
Assuming V0x = 0, then Vx ≠ 11/5 m/s.
 
  • Like
Likes 1 person
lewando said:
Assuming V0x = 0, then Vx ≠ 11/5 m/s.

yes, V0x=0 because it starts at rest. But how do you know the velocity is not 11/5 m/s then?
 
11/5 m/s is the average velocity. Vx is the instantaneous velocity at the 5m point.
 
  • Like
Likes 1 person
lewando said:
11/5 m/s is the average velocity. Vx is the instantaneous velocity at the 5m point.

Ohhhh.
And because of that, I cannot use any of the formulas which includes the Vx therefore the only correct equation for my current problem is the 3rd, correct?
 
Yes! :smile:
 
  • Like
Likes 1 person
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 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
4K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K