Looking for a pattern is not easy,how do you know when to stop looking and conclude there is no formula?
I think most series including reciprocals have no formulae,but some have and is it possible to check?
T=6g and T1=6g+4g=10g
But this is not the answer I had been taught,it is 2g.
how can it be when some other forces like 4g and T1 are pulling T?
Same question for m1.
Things on the one side of the pulley have no effects on the other?
An air balloon of mass M descending down with acceleration 'a'.A mass m is dropped out of it and the acceleration becomes a upwards.Find m in terms of M,a...
The right answer is 2Ma/(g+a)
The explanation I was given:
The mass out of the balloon now has a+g acceleration.
(How could it be when g...
Very very nice trick,impressive!
using this,i can make any eq with only degree as 1/2 a polynomial no matter,how many terms there are...
I'll be soon bringing another question about rational degrees(maybe) and I think the general question can also be solved by induction.Right?
Let me see.
No,I was thinking about A,B,C...representing polynomial functions on a variable (say x) but RHS a constant.There you can't do these kind of techniques when your purpose is to solve for x.
But nice (funny?) answer,for my less informative post.
There is a Q,derive the eqns of motion using calculus
and my txtbook derived
x=x0+v0t+1/2at sqr
and after all 3 eqns there is a note:
These eqns can be used for non uniformly accelerated motion too.
Now is my textbook wrong?
I think by a he means f(t) for instantaneous acceleration at t.
If...
no,my txtbook deroved the same equation taking a=dv/dt,integrating b/w the limits u and v
gives the 1st eq of motion
taking v=dx/dt and integrating and subbing v gives the 2 nd eq as x=ut+1/2at^2
but ther's a difference,a is instantaneous acceleration