A particle moves rectilinearly. Its displacement x at time t is given by x^2 = a t^2+ b where a and b are constants. Its acceleration at time t is proportional to
Answer (1)
Student Expert
31st Oct, 2019
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It is given that the displacement is :-
x^2 = at^2 + b.
On differentiating it with respect to time we get :-
2xdx = 2atdt + 0
On rearranging the terms we have :-
dx/dt = at/x (dx/dt is velocity) ........... ( 1)
Now x from the given equation can be written as x = (at^2 + b)
On substituting value of x from above to equation number 1 we get :-
dx/dt = at/(at^2 + b) ..............( 2)
Now, once again differentiating equation 2 with respect to time (derivative of velocity is acceleration) we get :-
dv/dt = A( acceleration )
{ (at^2 +b)*a - a^2*t^2/(at^2 +b) }/ (at^2 + b)
Therefore , A = ab/[(at^2 + b)^3/2] ......... ( 3)
Now since x = (at^2 + b) ...(4)
From eqns 3& 4, acceleration is proportional to 1/x^3
Thankyou
Greetings!
It is given that the displacement is :-
x^2 = at^2 + b.
On differentiating it with respect to time we get :-
2xdx = 2atdt + 0
On rearranging the terms we have :-
dx/dt = at/x (dx/dt is velocity) ........... ( 1)
Now x from the given equation can be written as x = (at^2 + b)
On substituting value of x from above to equation number 1 we get :-
dx/dt = at/(at^2 + b) ..............( 2)
Now, once again differentiating equation 2 with respect to time (derivative of velocity is acceleration) we get :-
dv/dt = A( acceleration )
{ (at^2 +b)*a - a^2*t^2/(at^2 +b) }/ (at^2 + b)
Therefore , A = ab/[(at^2 + b)^3/2] ......... ( 3)
Now since x = (at^2 + b) ...(4)
From eqns 3& 4, acceleration is proportional to 1/x^3
Thankyou
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