A particles velocity is given by the function :
v=kt^2 (v= k(t-squared)

where k is a constant and t is in seconds.
The particle's position at t=0, is -9 m. At t =3, the position is 9m. Determine the value of the constant k, and be sure to include the proper units.

Question

A particles  velocity is given by the function :
          v=kt^2    (v= k(t-squared)

where k is a constant  and t is in seconds. 
The particle's position at t=0, is -9 m. At t =3, the position is 9m. Determine the value of the constant k, and be sure to include the proper units.

fifciol · Answer

start from this equation :
$$v=\frac{ dx }{ dt }\Rightarrow vdt=dx\Rightarrow \int vdt=\int dx \Rightarrow x=\int vdt$$

fifciol · Answer

substitute v=kt^2 and solve that integral
$$x=\int\limits kt^2dt=\frac{ 1 }{ 3 }kt^3+C$$

fifciol · Answer

then substitute :
-9= 1/3 *k*0^3 +C
9=1/3*k*3^3+C
so
from 1-st equation you can immidiately notice that:
C=-9
and by substituting that to 2-nd equation you'll get k=2

fifciol · Answer

since v=kt^2
k=v/t^2
so the unit of k is m/s/s^2=m/s^3

anonymous · Answer

thanx @fifciol, sorry I wasn't online for a step by step tutorial