a particle moves under the equation t=px^2+qx where p and q are positive constant. Determine the retardation of the particle where t is time and x is position
Answers
Answer:
It's too simple dear.... You must have solved it yourself...
Explanation:
retardation is second derivative of position with respect to time....
Your function is in x that is t as a function of x is given.... Convert it first to x as a function of t.... Or do in this way....
t=px^2+qx
Taking time derivative....
dt=(2px + q) dx
dx / dt = v=1/(2px + q)
= (2px + q) ^-1
Again taking time derivative ...
d^2x/dt^2 = acc = [ {-1/(2px+q)^2}*2p ]....
a= -( v^2)*2p....
Alternative.....
Express x as a function of time t...
t = px^2 +qx
t = p( x^2 +(q/p)x)
t/p = x^2 +(q/p) x
t/p = [(x +(q/2p))]^2 - (q/2p)^2...
t/p + (q/2p)^2 = [(x +(q/2p))]^2
[(t*4p+q^2)/4p^2] = [(x +(q/2p))]^2..
[(x +(q/2p))] = {[(t*4p+q^2)/4p^2]}^ 1/2.....
x = ( {[(t*4p+q^2)/4p^2]}^ 1/2) - (q/2p)....
Take two times time derivative of x... You will get acceleration in terms of time t....
I'm not going to do that. .. It's a better way of learning to do something by yourself.... I helped you in expressing x as a function of time t....
Now complete it soon and ask me if you have any problem.....
Answered by an IIT JEE ASPIRANT.... See my profile dear....