Calculating acceleration using Newton's 3rd law

  • Thread starter Thread starter timelesstrix0
  • Start date Start date
  • Tags Tags
    Acceleration Law
AI Thread Summary
The discussion centers around calculating Jimmy's speed after he pushes Drake for 0.50 seconds, with Drake coming to rest 25 meters away after 10 seconds. Participants highlight the importance of understanding how the push duration affects momentum and acceleration calculations, particularly in relation to Newton's Third Law. There is confusion regarding the role of friction, as it is initially stated to be ignored, yet it is necessary for determining Drake's deceleration. The average velocity formula is noted to apply to the entire distance, not just the initial or final velocities, and the SUVAT equations are suggested for constant acceleration scenarios. Ultimately, the push duration is relevant for calculating the velocities of both individuals immediately after the push.
timelesstrix0
Messages
5
Reaction score
0
Member warned about posting without the homework template
So i have a challenging question given by my teacher:
Jimmy and Drake are at rest and on ice (therefore, we can ignore friction). Jimmy pushes
Drake for 0.50 seconds. Drake comes to rest 25 metres away from where they were at rest
10.0 seconds after the push.
Jimmy has a mass of 60 kg and Drake has a mass of 75 kg. How fast will Jimmy be going
after he has finished pushing Drake?!

i wana know how the "0.50 seconds" of push time will affect how i do the question...
Right now i can find Drake's acceleration and use that to find the force applied on Drake and use 3rd law and say that Jimmy will have the same amount of force... but idk if I am doing this right because i do not know how the 0.5 seconds of push time is involved in the question
for my velocity calculation (which will be used for the acceleration) will i have to use 0.5 seconds as the initial time value when doing
v = Δd/Δt
so it will be like: v = 25-0/10-0.5
 
Physics news on Phys.org
If "we can ignore friction", what causes Drake to come to "rest 25 metres away"?
 
timelesstrix0 said:
i wana know how the "0.50 seconds" of push time will affect how i do the question...

this part is very important -as the things started with it...

in this time interval Jimmy was pushing the drake ,so a force was being applied by J on D ; and Newtons Third law operates to say that The same force must have been applied by Drake on Jimmy for the same time interval - this leads to change in momenta of the two people

so Force .Time interval = change in momentum if the time involved is small and then you can proceed further.
 
Please double check you have stated the question accurately. As it stands, it makes no sense. If there is no friction, Drake will not come to rest. To figure out Jimmy's speed just after the push we need to find Drake's, and to find Drake's speed from the distance he takes to come to rest we need to know the coefficient of friction.
 
  • Like
Likes drvrm
no i stated the question as it was given by my teacher (i copy pasted from the word document that my teacher gave us)
 
timelesstrix0 said:
no i stated the question as it was given by my teacher (i copy pasted from the word document that my teacher gave us)
I overlooked that we know how long Drake takes to come to rest, so we can determine the coefficient of friction. All that is wrong with the question is that it should not say friction should be ignored.
The equation v=Δd/Δt gives the average velocity, not the initial or final velocity.
Are you familiar with the SUVAT equations? These apply when acceleration is constant, which it is here.
Also, note that the 10 seconds starts after the push ends, so do not subtract 0.5 from it.
 
haruspex said:
I overlooked that we know how long Drake takes to come to rest, so we can determine the coefficient of friction. All that is wrong with the question is that it should not say friction should be ignored.
The equation v=Δd/Δt gives the average velocity, not the initial or final velocity.
Are you familiar with the SUVAT equations? These apply when acceleration is constant, which it is here.
Also, note that the 10 seconds starts after the push ends, so do not subtract 0.5 from it.
so does the 0.5 seconds have a purpose when I am doing my calculations??
 
timelesstrix0 said:
no i stated the question as it was given by my teacher (i copy pasted from the word document that my teacher gave us)
the reason for stopping velocity going to zero must be there
 
timelesstrix0 said:
so does the 0.5 seconds have a purpose when I am doing my calculations??
Not that I can see.
 
  • #10
timelesstrix0 said:
so does the 0.5 seconds have a purpose when I am doing my calculations??

this time interval is the time during which pushing and reaction was being applied so this interval can be used in calculating the velocity acquired by Drake after which he traveled further and moreover the velocity acqired by jimmy can also be calculated .
one can equate the momenta also as the same force was being applied .
 
  • #11
drvrm said:
this time interval is the time during which pushing and reaction was being applied so this interval can be used in calculating the velocity acquired by Drake after which he traveled further and moreover the velocity acqired by jimmy can also be calculated .
one can equate the momenta also as the same force was being applied .
Again, it comes back to this issue of friction. The only reason that the duration of the push may matter is if there is a net external horizontal force acting during that period. If not, momentum is conserved during the push, and we can derive Jimmy's speed at end of push directly from Drake's. Since the two have different masses, the frictional force is different, so there is a small net horizontal force. I suspect the "friction can be ignored" clause was meant to apply to the duration of the push.
 
Back
Top