SOLUTION: The acceleration of a particle at a position x on the number line is given bya=-25x. Initially the particle is at the origin and has a velocity of 10m/s. Find the velocity after

Consider a harmonic oscillation


    x = a*sin(bt)      (1)


with the unknown amplitude " a " and unknown coefficient " b ".


For such oscillations,  the velocity is  


    v = x'(t) = ab*cos(bt)


and the acceleration is 


    a = x''(t) = -ab^2*sin(bt).


Also notice that for this harmonic motion the initial position is  x = 0  at t= 0.


In your case, take  ab = 10  and ab^2 = -25.


Then you will have EXACTLY the described motion.


From (2),  you have first  


    b =  =  = -2.5,

and then

    a =  =  = -4.


So, your motion is


    x = -4*sin(-2.5*t) = 4*sin(2.5t).

with

    v = x'(t) = 4*2.5*cos(2.5t) =  =  =  = 8.66.


ANSWER.  The velocity after   seconds is 8.66 m/s.

    a)  Either to know from Math that the solution/solutions to the equation


            x''(t) = -ax


         are the functions  a*sin(bt),  a*cos(bt),    or



    b)  to know from Physics, that the motion where the returning force is proportional to the displacement from the equilibrium 

        taken with the opposite sign, is a harmonic motion.