Answer:
[tex]v(t)=2.7*e^{0.5kt}\\\\ v(s)=\frac{k}{2}*s+2.7-1.3k\\\\[/tex]
For t=2.7s and k=0.3 m/s:
[tex]v(2.7s)=1.80m/s[/tex]
For s=6m and k=0.3 m/s:
[tex]v(6m)=6.69m/s\\\\[/tex]
Explanation:
Definition of acceleration:
[tex]a=\frac{dv}{dt} =0.5kv[/tex]
we integrate in order to find v(t):
[tex]\frac{dv}{v} =-0.5kdt[/tex]
[tex]\int\limits^v_0 { \frac{dv}{v}} \, =-0.5k\int\limits^t_0 {dt} \,[/tex]
[tex]ln(v)=-0.5kt+C\\\\ v=A*e^{-0.5kt}[/tex] A=constant
Definition of velocity:
[tex]v=\frac{ds}{dt} =A*e^{-0.5kt}[/tex]
We integrate:
[tex]v=\frac{ds}{dt} \\s=- (2/k)*A*e^{-0.5kt}+B[/tex] B=constant
But:
[tex]v=A*e^{-0.5kt}[/tex]⇒[tex]s= -(2/k)*v+B[/tex]
[tex]v(s)=-(\frac{k}{2} )(s-B)=D-\frac{k}{2}*s[/tex] D=other constant
Initial conditions:t = 0 are s0 = 2.6 m and v0 = 2.7 m/sec:
[tex]v(t)=A*e^{-0.5kt}\\ 2.7=Ae^{-0.5k*0}\\ 2.7=A\\[/tex]
[tex]v(s)=D-\frac{k}{2}*s\\2.7=D-\frac{k}{2}*2.6\\D=2.7+1.3k[/tex]
So:
[tex]v(t)=2.7*e^{-0.5kt}\\\\ v(s)=\frac{k}{2}*s+2.7+1.3k\\\\[/tex]
For t=2.7s and k=0.3 m/s:
[tex]v(2.7s)=2.7*e^{-0.5*0.3*2.7}=1.80m/s[/tex]
For s=6m and k=0.3 m/s:
[tex]v(6m)=\frac{0.3}{2}*6+2.7+1.3*0.3=6.69m/s\\\\[/tex]
