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

[cneedshelp]

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?

 

[1MileCrash]

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)

 

[cneedshelp]

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?

 

[1MileCrash]

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.

 

[cneedshelp]

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?

 

[1MileCrash]

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.