Answer:
(a)The position of the particle after a time t is
[tex]S(t)=\frac{t^3}3+9t+c[/tex]
(b)The position of the particle after a time t is
[tex]S(t)=\frac{t^3}3+9t+3[/tex]
Step-by-step explanation:
We know that, the first order derivative of the position of an object is the velocity of the object.
(a)
Given that, the velocity of a particle moving in straight line is
[tex]V(t)=t^2+9[/tex]
[tex]\Rightarrow \frac{dS(t)}{dt}=t^2+9[/tex]
[tex]\Rightarrow {dS(t)}=t^2dt+9\ dt[/tex]
Integrating both sides
[tex]\int {dS(t)}=\int t^2dt+\int9\ dt[/tex]
[tex]\Rightarrow S(t)=\frac{t^3}3+9t+c[/tex] [ c is an arbitrary]
The position of the particle after a time t is
[tex]S(t)=\frac{t^3}3+9t+c[/tex]
(b)
Given that S= 3 at time t=0
[tex]\therefore 3=S(t)=\frac{0^3}3+9.0+c[/tex]
[tex]\Rightarrow c=3[/tex]
The position of the particle after a time t is
[tex]S(t)=\frac{t^3}3+9t+3[/tex]
