document.write( "Question 356762: An unidentified object moves along the s-axis, with displacement s = s(t) (meters), velocity v = v(t) (m/sec), and acceleration a = a(t) (m/sē). It so happens that the velocity and displacement are related by the equation v = sqrt(8s+16). Moreover, at the instant t = 0, the object is observed at s = 6. A) Show that a is constant, and show its value. B) Graph v as a function of s. C) Graph v as a function of t. \n" ); document.write( "
Algebra.Com's Answer #254669 by Jk22(389)\"\" \"About 
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v=Sqrt(8s+16),
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\n" ); document.write( "A) a=dv/dt=1/2*1/Sqrt(8s+16)*8v=4v/Sqrt(8s+16)=4v/v=4 (m/s^2)
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\n" ); document.write( " (Other way : v^2*m/2=Kinetic Energy=m*(4s+8)
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\n" ); document.write( " Potential Energy = Constant - Kinetic Energy = Const - 4ms -8m
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\n" ); document.write( " Force = - d(Potential Energy)/ds = 4m = m*a => a=4 m/s^2)
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\n" ); document.write( "B) v=Sqrt(8s+16)
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\n" ); document.write( "with v=ds/dt we get a differential equation :
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\n" ); document.write( "ds/Sqrt(8s+16)=dt=ds/(2Sqrt(2s+4))
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\n" ); document.write( "Integration gives :
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\n" ); document.write( " t+C=1/2*Sqrt(2s+4),
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\n" ); document.write( " indeed d(t+C)=dt, C is a constant
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\n" ); document.write( " 1/2*d(Sqrt(2s+4))=1/2*1/(2Sqrt(2s+4))*2=1/(2*Sqrt(2s+4))=1/Sqrt(8s+16)
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\n" ); document.write( " s(t=0)=6 : at t=0, s=6, in the equation : C=1/2*Sqrt(12+4)=1/2*Sqrt(16)=2
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\n" ); document.write( "the function displacement towards time is :
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\n" ); document.write( " s(t)=1/2*(2t+4)^2-4
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\n" ); document.write( " =1/2(4t^2+16*t+16)-16
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\n" ); document.write( " =2t^2+8t-8 (m)
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\n" ); document.write( "C) the velocity is : v(t)=ds/dt=
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\n" ); document.write( " =4t+8 (m/s)
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\n" ); document.write( "Verification : acceleration is : a(t)=dv/dt=4 (m/s^2)
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