#kinematics as easy
103 messages · Page 1 of 1 (latest)
s=ut+1/2at²
find a
then use v=u+at
as it moves to A, it will have negative displacement
but since your acceleration will also be negative, it'll be positive
@shy stag
can u write the whole thing out for me 🙏 im super confuzzled
you move to B
how else are you gonna find the velocity at A
Well can i show u what i initially did
sure
,rotate
yeah you can't do this
why
the velocity at B is your initial velocity for your A calculation
but you're trying to find the velocity as it passes A
that's Va
Vb=Ua
What so because the acceleration is constant we find that first as it will be the same throughout and then find the initial velocity
you already know the initial velocity
initial velocity = velocity at B
you dont know the final velocity, which is the velocity at A
for these sorts of problems you have to do this every time
the question said as it passed A
so if we worked backwards, say from B to A
we already know what the velocity was at B
we don't know the velocity at A
and we can find the acceleration
because we have the distance
I’m actually mega confused maybe bc it’s late
Can i send u another
How do u do part a @languid agate
what's the displacement at rest?
0
you can now do (a)
you can't have negative time
you will probably get a positive and negative answer
yes
long day
I found ur issue
you have a double negative
0=12t-1/2(3)t²
(3/2)t²-12t=0
t(3/2t-12)
t=0 or t=12×2/3
t=8
how is it 4
you have to...or else you get imaginary roots
if both are positive
you essentially get
Huh
(3/2)t²+12t
U didn’t use -3
I did
negative sign is on the outside
I can write it 0=12t+1/2(-3)t² if you want
Wait so 12 is u not v?
Ohh
so 12 is the initial velocity
yeah ive been stupid
yeah rest assumes v=0
been a long day
yeah im an idiot you can't even use a quadratic