Freefall Problems: Acceleration at 2s, 10s & Any t

  • Thread starter Thread starter cneedshelp
  • Start date Start date
  • Tags Tags
    Freefall
Click For Summary

Homework Help Overview

The discussion revolves around the acceleration of a freely falling object dropped from rest, specifically at different time intervals: the end of the 2nd second, the end of the 10th second, and at any elapsed time t. The subject area is kinematics, focusing on the effects of gravity on falling objects.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between velocity and acceleration, questioning the sign convention used for acceleration due to gravity. Some participants attempt to clarify whether acceleration should be considered negative or positive based on the chosen coordinate system.

Discussion Status

The discussion is active, with participants providing differing perspectives on the sign of acceleration and velocity. Some guidance is offered regarding the agreement of signs in a consistent coordinate system, but no consensus has been reached on the interpretation of the acceleration value.

Contextual Notes

There is mention of a teacher's input regarding the negative value of acceleration due to gravity, which adds to the complexity of the discussion. Participants are navigating the implications of different coordinate systems and their effects on the interpretation of motion.

cneedshelp
Messages
14
Reaction score
0
For a freely falling object dropped from rest, what is the acceleration at the following times?
a.the end of the 2nd second
b. the end of the 10th second
c. the end of any elapsed time t

Given-
A= -9.80m/s^2
Vi=0m/s

Do you have to find the final velocity, then acceleration?
 
Physics news on Phys.org
Your initial velocity is 0 m/s. The object is falling, but there is no reason to attribute a negative value to acceleration due to gravity on this coordinate system. Velocity increases by 9.8 m/s every second.

Therefore,
v = 9.8(t)

(I can only assume your question asks for VELOCITY after elapsed time, as acceleration is constant)
 
1MileCrash said:
Your initial velocity is 0 m/s. The object is falling, but there is no reason to attribute a negative value to acceleration due to gravity on this coordinate system. It accelerates by 9.8 m/s every second.

v = 9.8(t)

Why wouldn't it be negative, I thought when it said dropped from rest meant it is lower than 0?
 
If it is dropped from rest, it is falling towards earth. Acceleration is in that same direction, therefore the signs of velocity and acceleration must agree.
 
1MileCrash said:
If it is dropped from rest, it is falling towards earth. Acceleration is in that same direction, therefore the signs of velocity and acceleration must agree.

I asked my teacher about it, and she said the 9.8 is always negative, so how do i work out the problems?
 
No, it's not. It's negative if you attribute a coordinate system in which it is negative. In your example, if you do so, you must also attritube your velocity as negative since they are in the same direction (as I noted above)

- v = -9.8(t)
and
v = 9.8(t)

are the exact same thing.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
34
Views
3K
  • · Replies 14 ·
Replies
14
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K