Answer:
[tex]s(t) = t^{\frac{3}{2} } +6[/tex]
Step-by-step explanation:
given that a particle is moving with velocity given by the function v(t)
We have to find the position of the particle
[tex]v(t) = 1.5\sqrt{t}[/tex]
Since v is the derivative of s, we can find s by integrating v wrt t
[tex]s(t) = \int v(t) dt \\=\int 1.5\sqrt{t} dt\\=1.5 *\frac{2}{3} t^{\frac{3}{2} } +C[/tex]
where C is the arbitrary constant of integration.
Use the fact s(4) = 14
[tex]s(t) = 1.5 *\frac{2}{3}* t^{\frac{3}{2} } +C\\s(4) = 1.5 *\frac{2}{3}*4^{\frac{3}{2} } +C=14\\8+C=14\\C = 6[/tex]
So position of particle at time t is given by
[tex]s(t) = t^{\frac{3}{2} } +6[/tex]
