equations of motion (1 Viewer)

[kkk579]

So usually these qs branch out into a few types of qs but i have qs about two of them.

1)Qs where an object is falling to the ground from a certain height:
Must the initial velocity always be 0? Can u put the final velocity at 0 cos technically when it hits the ground and stops moving it’s velocity is 0.

2)Qs where an object is thrown into the air:
Can you put the final velocity at 0? Since at a very minuscule point in time it stops moving while at its peak height.

I generally use the sign convention that going up is positive and going down is negative so wouldn’t ag literally always equal -9.8 not 9.8?

 

[kkk579]

Here is an example of how i usually do the qs:
Is this the correct or wrong way of doing it?

The q is: A stone is dropped into a mine shaft and reaches the bottom in 4 sec. How deep is the shaft?

 

[carrotsss]

you can’t say that the ending velocity is zero just because of that split moment, the equation of motion is modelling the motion until that happens and doesn’t really know when the bottom actually is

however for the peak height you can use that, but you have to remember that it’s only half of the time taken

and yeah gravity is pretty much always negative but you do have to kinda understand what the situation is (ie if something is launched downwards then gravity is adding to it)

 

[kkk579]

you can’t say that the ending velocity is zero just because of that split moment, the equation of motion is modelling the motion until that happens and doesn’t really know when the bottom actually is

ohhh okay, it’s weird tho cos when i’m calculating time and stuff and use ag=-9.8 it gives me a negative number. i think my understand if sign conventions and when ag is - or + is very blurred. do u mind explaining?

 

[carrotsss]

ohhh okay, it’s weird tho cos when i’m calculating time and stuff and use ag=-9.8 it gives me a negative number. i think my understand if sign conventions and when ag is - or + is very blurred. do u mind explaining?

sorry I edited my post to answer that stuff after

 

[kkk579]

you can’t say that the ending velocity is zero just because of that split moment, the equation of motion is modelling the motion until that happens and doesn’t really know when the bottom actually is

however for the peak height you can use that, but you have to remember that it’s only half of the time taken

and yeah gravity is pretty much always negative but you do have to kinda understand what the situation is (ie if something is launched downwards then gravity is adding to it)

when is gravity not negative?

 

[carrotsss]

when is gravity not negative?

gravity is not negative when you’re saying that your velocity is positive, and it is in the downwards direction

or you can avoid this and just always remember to use downwards velocities as negative

 

[kkk579]

gravity is not negative when you’re saying that your velocity is positive, and it is in the downwards direction

or you can avoid this and just always remember to use downwards velocities as negative

when an object is moving downwards but u get ur final velocity as a positive number, do u just put the negative in front of it although it’s not what u calculated as it’s moving downwards?

 

[carrotsss]

when an object is moving downwards but u get ur final velocity as a positive number, do u just put the negative in front of it although it’s not what u calculated as it’s moving downwards?

yeah pretty much, or usually you’ll just say that it’s downwards after the velocity (velocity is a vector quantity so you must specify). it mostly doesn’t matter much, you just kinda use common sense to figure out the sign of the final answer (like obvs ur time isn’t gonna be negative)

 

[kkk579]

when an object is moving downwards but u get ur final velocity as a positive number, do u just put the negative in front of it although it’s not what u calculated as it’s moving downwards?

oh what i didn’t know u could do that

 

[kkk579]

yeah pretty much, or usually you’ll just say that it’s downwards after the velocity (velocity is a vector quantity so you must specify). it mostly doesn’t matter much, you just kinda use common sense to figure out the sign of the final answer (like obvs ur time isn’t gonna be negative)

yeah that makes sense, tysm

so let’s say upwards is positive and downwards direction is negative. if an object is moving upwards then ag=-9.8. if an object is moving downwards ag is still -9.8?

 

[carrotsss]

yeah that makes sense, tysm

so let’s say upwards is positive and downwards direction is negative. if an object is moving upwards then ag=-9.8. if an object is moving downwards ag is still -9.8?

yep

 

[kkk579]

yep

tysm this cleared up my misunderstandings a lot

 

[kkk579]

you can’t say that the ending velocity is zero just because of that split moment, the equation of motion is modelling the motion until that happens and doesn’t really know when the bottom actually is

however for the peak height you can use that, but you have to remember that it’s only half of the time taken

and yeah gravity is pretty much always negative but you do have to kinda understand what the situation is (ie if something is launched downwards then gravity is adding to it)

also what did u mean by “only half of the time taken”

 

[kkk579]

1/2 total time = peak because it's a parabola and when v = 0

oh so like the amount of time to reach the peak height is half the entire time since the entire time includes the object coming back down to the ground

 

[kkk579]

How about when an object is moving upwards into the air and comes back down? Do we say u=0m/s or v=0m/s?

 

[kkk579]

gravity is not negative when you’re saying that your velocity is positive, and it is in the downwards direction

or you can avoid this and just always remember to use downwards velocities as negative

gravity would also not be negative when ur velocity is negative in the upwards direction as well right?

 

[Eagle Mum]

How about when an object is moving upwards into the air and comes back down? Do we say u=0m/s or v=0m/s?

In this scenario, neither u nor v are 0 m/s. If an object is stationary (initial velocity of 0 m/s), it will not move upwards on its own accord; the only force acting on it is gravity which is in a downward direction.

 

[kkk579]

In this scenario, neither u nor v are 0 m/s. If an object is stationary (initial velocity of 0 m/s), it will not move upwards on its own accord; the only force acting on it is gravity which is in a downward direction.

Oh okay thanks

 

[carrotsss]

Oh okay thanks. The q would typically either give us u or v for these type of qs then?

yes