Question

He acceleration of particle moving along x-axs is given by a=3x+2 where x is the position at any time t. the position of the particle at t=0 is x=2m. assume the particle's velocity to be zero at t=0. find the velocity of particle as a function of x

Answer 1

Answer: $v = 3xt + 2t$

Explanation: acceleration = 3x+2

x = position of particle at time t. the position of particle at t=0 is x=2m

now acceleration is rate of change of velocity

therefore a = dv/dt

=> dv = adt

integrate on both sides

=> $\int dv = a\int dt \\\\ \int dv =\int (3x+2)dt\\\\ v = 3xt + 2t$

Also:

$v^{2} - u^{2} = 2ax$

where u is initial velocity, which is zero and v is final velocity

=>

$v^{2} - 0 = 2(3x+2)x$

=> $v^{2} = 2(3x+2)x$

=> $v=\sqrt{6x^{2} + 4x }$