Answer:
acceleration of the particle as a function of time is given as
[tex]a(t)=-\frac{ks}{t}[/tex]
and position of the particle as a function of time is given as
[tex]x(t)=ut+\frac{at^2}{2} \\x(t)=V_{0}t-\frac{\frac{ks(t)}{t}t^2}{2}[/tex]
[tex]x(t)=V_{0}t-\frac{kst}{2}[/tex]
Explanation:
we have given,
[tex]v=v_{o}− ks[/tex]
at t=0, s=0
therefore initial velocity,u
u=v=[tex]V_{o}[/tex]
v=u+at.........(1)
we know from first equation of motion
let acceleration at time t be a(t), and position be x(t)
put value of v,u and a in equation 1
we got,
[tex]V_{o}-ks=V_{o}+a(t)t[/tex]
Therefore,acceleration of the particle as a function of time is given as
[tex]a(t)=-\frac{ks}{t}[/tex]
and position of the particle as a function of time is given as
[tex]x(t)=ut+\frac{at^2}{2} \\x(t)=V_{0}t-\frac{\frac{ks}{t}t^2}{2}[/tex]
[tex]x(t)=V_{0}t-\frac{kst}{2}[/tex]
