If the displacement of a body is given by `S = at^2 + bt + c` and a, b, c are constants then show that if v be the velocity of ‘the particle then--------`4a(s -c) = v^2 - b^2`.

pv^a=c (a and c are constants) then show that v^2(d^2p)/(dv^2)=a(a+1)p .

If a,b and c are constant of motion and s = at^(2)+bt+ c , then prove that 4a(s-c) = v^(2) - b^(2) .

The displacement x of a particle in rectilinear motion at time t is represented as x^(2) = at^(2) +2bt+c , where a ,b, and c are constants. Show that the acceleration of this particle varies (1)/(x^(3)) .

The displacement of a particle is given by y = a +bt+ct^(2)+dt^(4) . The initial velocity and acceleration are respectively

Displacement (x) and time (t) of a particle in motion are related as x = at + bt^(2) -ct^(3) where a,b, and c, are constants. Velocity of the particle when its acceleration becomes zero is

If a, b, c are In A.P., then show that, (a+2b-c) (2b+c-a)(c+a-b) = 4abc

If pv^(a)=c (a and c are costants) then show that v^(2)(d^(2)p)/(dv^(2))=a(a+1)p

The displacement (s ) of a particle depends on time (t) as s = 2at^(2)-bt^(3) . Then

A particle starts with the velocity u and moves in a straight line, its acceleration being always equal to its displacement. If v be the velocity when its displacement is x, then show that v^2= u^2+x^2 .

If the velocity of a particle is v = At+Bt^(2) , where A and B are constant then the distance travelled by it between 1 s and 2 s is - (A) (3A +7B) (B) (3)/(2)A+(7)/(3)B (C) (A)/(2)+(B)/(3) (D) (3)/(2)A+4B