Question

position x of a particle at any instant is related with velocity as v=(2x+9)½. The particle starts from origin. Then initial acceleration and velocity​

Answer 1

Answer:

solved

Explanation:

v = (2x+9)½ = dx/dt

(2x+9)-½ .dx = dt

∫(2x+9)-½ .dx = ∫dt

(2x+9)½ = t + c

particle starts from origin , x=0 at t=0  , c = 3

(2x+9)½ = t + 3

2x+9 = (t + 3)² = t² + 6t +9

2x =  t² + 6t

x = t²/2  + 3t

v  = t +   3 , Vi = 3 m/s

a = 1 m/s²