IIT JEE , AIEEE Forum - Physics , Mechanics

kislay

pls solve

pranil_2007

a=(25+200(2+2.45))/30

kislay

HEY NO IS TRYING??????????????/
:-(

pjd_0501

is the ans.......-12.7m/s²

bhupesh

Dear
As per your question the mathematical equation describing such type of motion will be

v = -kx + c

v= velocity
x = displacement

using the condition provide in your question you can find the values of k and c

for x= 0 v= 20 which gives c= 20
similarly for x= 30 v= 20 which gives the value of k

so now you have the equation differentiate this you will find the acceleration in terms of velocity as the above expression gives the value of v you can easily find the answer and can also prove it
x= dis

kislay

SIR I HAVE GOT ANSWER TO FIRST PART ,BUT AM NOT ABLE TO PROVE.PLS HELP

bhupesh

Dear
Its easy what you have to do is solve the differential equation
dx/dt = -kx + c
You need to integrate it right ?

So put lower limit for t =0 to t= t and for x=s to x= 30 ( x=s because it is possible for the object to be at some position when t= 0 )

So when you will try to integrate it you will find something interesting which prohibits the object to reach the x= 30

There are of course other ways to solve

Ask & Discuss Questions with Community & Experts