Question

The velocity of a particle moving in a straight line is given by v(t) = t2 + 9. (a) Find an expression for the position s after a time t . S(t) = + C (b) Given that s = 3 at time t = 0, find the constant of integration C. C = Find an expression for s in terms of t without any unknown constants.

Answer 1

Answer:

(a)The position of the particle after a time t is

$S(t)=\frac{t^3}3+9t+c$

(b)The position of the particle after a time t is

$S(t)=\frac{t^3}3+9t+3$

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

$V(t)=t^2+9$

$\Rightarrow \frac{dS(t)}{dt}=t^2+9$

$\Rightarrow {dS(t)}=t^2dt+9\ dt$

Integrating both sides

$\int {dS(t)}=\int t^2dt+\int9\ dt$

$\Rightarrow S(t)=\frac{t^3}3+9t+c$  [ c is an arbitrary]

The position of the particle after a time t is

$S(t)=\frac{t^3}3+9t+c$

(b)

Given that S= 3 at time t=0

$\therefore 3=S(t)=\frac{0^3}3+9.0+c$

$\Rightarrow c=3$

The position of the particle after a time t is

$S(t)=\frac{t^3}3+9t+3$