Question 356762
 v=Sqrt(8s+16), 
 
A) a=dv/dt=1/2*1/Sqrt(8s+16)*8v=4v/Sqrt(8s+16)=4v/v=4 (m/s^2)
 
 (Other way : v^2*m/2=Kinetic Energy=m*(4s+8)
 
 Potential Energy = Constant - Kinetic Energy = Const - 4ms -8m
 
 Force = - d(Potential Energy)/ds = 4m = m*a => a=4 m/s^2)
 
 
 
 
B) v=Sqrt(8s+16)
 {{{ graph( 300, 200, 0, 5, 0, 10, sqrt(8x+16) ) }}}
 
 
 
with v=ds/dt we get a differential equation : 
 
ds/Sqrt(8s+16)=dt=ds/(2Sqrt(2s+4))
 
Integration gives : 
 
 t+C=1/2*Sqrt(2s+4), 
 
 indeed d(t+C)=dt, C is a constant
 
 1/2*d(Sqrt(2s+4))=1/2*1/(2Sqrt(2s+4))*2=1/(2*Sqrt(2s+4))=1/Sqrt(8s+16)
 
 
 s(t=0)=6 : at t=0, s=6, in the equation : C=1/2*Sqrt(12+4)=1/2*Sqrt(16)=2
 
the function displacement towards time is : 
 
 s(t)=1/2*(2t+4)^2-4
 
   =1/2(4t^2+16*t+16)-16
 
   =2t^2+8t-8 (m)
 
 
C) the velocity is :  v(t)=ds/dt= 
 
   =4t+8 (m/s)
 
{{{ graph( 300, 200, 0, 5, 0, 40, 4*x+8 ) }}}
 
 
 
Verification : acceleration is : a(t)=dv/dt=4 (m/s^2)