Answer with Explanation:
By definition of acceleration we have
[tex]a=\frac{dv}{dt}[/tex]
Given [tex]a=kv^{2}[/tex], using this value in the above equation we get
[tex]kv^2=\frac{dv}{dt}\\\\\frac{dv}{v^{1}}=kdt[/tex]
Upon integrating on both sides we get
[tex]\int \frac{dv}{v^2}=\int kdt\\\\-\frac{1}{v}=kt+c[/tex]
'c' is the constant of integration whose value can be found out by putting value of 't' = 0 and noting V =4 m/s
Thus [tex]c=-\frac{1}{4}[/tex]
the value of 'k' can be found by using the fact that at t= 30 seconds velocity = 26 m/s
[tex]\frac{-1}{26}=k\times 30-\frac{1}{4}\\\\\therefore k=\frac{\frac{-1}{26}+\frac{1}{4}}{30}\\\\k=7.05\times 10^{-3}[/tex]
Hence the velocity as a function of time is given by
[tex]v(t)=\frac{-1}{7.05\times 10^{-3}t-\frac{1}{4}}[/tex]
By definition of velocity we have
[tex]v=\frac{dx}{dt}[/tex]
Making use of the obtained velocity function we get
[tex]\frac{dx}{dt}=\frac{-1}{kt-\frac{1}{4}}\\\\\int dx=\int \frac{-dt}{kt-\frac{1}{4}}\\\\x(t)=\frac{-1}{7.05\times 10^{-3}}\cdot ln(7.05\times 10^{-3}t-\frac{1}{4})+x_o[/tex]
here [tex]x_o[/tex] is the constant of integration
