# Calculating acceleration using Newton's 3rd law

1. Mar 29, 2016

### timelesstrix0

• 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 im 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

2. Mar 29, 2016

### HallsofIvy

Staff Emeritus
If "we can ignore friction", what causes Drake to come to "rest 25 metres away"?

3. Mar 29, 2016

### drvrm

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.

4. Mar 29, 2016

### haruspex

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.

5. Mar 29, 2016

### timelesstrix0

no i stated the question as it was given by my teacher (i copy pasted from the word document that my teacher gave us)

6. Mar 29, 2016

### haruspex

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.

7. Mar 29, 2016

### timelesstrix0

so does the 0.5 seconds have a purpose when im doing my calculations??

8. Mar 29, 2016

### drvrm

the reason for stopping velocity going to zero must be there

9. Mar 29, 2016

### haruspex

Not that I can see.

10. Mar 29, 2016

### drvrm

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 travelled 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. Mar 29, 2016

### haruspex

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.