Question

Find the position function of a particle moving along a coordinate line that satisfies the given condition v(t) = 6t2 - 4t + 5,s(1) = - 4 s(t) =

Answer 1

Answer:

$s(t) = 2t^{3} - 2t^{2} + 5t - 9$

Step-by-step explanation:

The position is the integrative of the velocity.

We have that:

$v(t) = 6t^{2} - 4t + 5$

So

$s(t) = \int {v(t)} \, dt$

$s(t) = \int {(6t^{2} - 4t + 5)} \, dt$

$s(t) = 2t^{3} - 2t^{2} + 5t + K$

In which k is the constant of integration, which we use the initial condition to find.

s(1) = -4. So

$-4 = 2*(1)^{3} - 2*(1)^{2} + 5(1) + K$

$K = -9$

So

$s(t) = 2t^{3} - 2t^{2} + 5t - 9$