Question

A particle is moving with the given data. Find the position of the particle. v(t) = 1.5√t , s(4) = 14.

Answer 1

Answer:

$s(t) = t^{\frac{3}{2} } +6$

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

$v(t) = 1.5\sqrt{t}$

Since v is the derivative of s, we can find s by integrating v wrt t

$s(t) = \int v(t) dt \\=\int 1.5\sqrt{t} dt\\=1.5 *\frac{2}{3} t^{\frac{3}{2} } +C$

where C is the arbitrary constant of integration.

Use the fact s(4) = 14

$s(t) = 1.5 *\frac{2}{3}* t^{\frac{3}{2} } +C\\s(4) = 1.5 *\frac{2}{3}*4^{\frac{3}{2} } +C=14\\8+C=14\\C = 6$

So position of particle at time t is given by

$s(t) = t^{\frac{3}{2} } +6$