Question

A particle is moving with the given data. Find the position of the particle. a(t) = 2t + 5, s(0) = 6, v(0) = −5

Answer 1

Answer:

The position of the particle is described by $s(t) = \frac{1}{3}\cdot t^{3} + \frac{5}{2}\cdot t^{2} - 5\cdot t + 6,\forall t \geq 0$

Step-by-step explanation:

The position function is obtained after integrating twice on acceleration function, which is:

$a(t) = 2\cdot t + 5$, $\forall t \geq 0$

Velocity

$v(t) = \int\limits^{t}_{0} {a(t)} \, dt$

$v(t) = \int\limits^{t}_{0} {(2\cdot t + 5)} \, dt$

$v(t) = 2\int\limits^{t}_{0} {t} \, dt + 5\int\limits^{t}_{0}\, dt$

$v(t) = t^{2}+5\cdot t + v(0)$

Where $v(0)$ is the initial velocity.

If $v(0) = -5$, the particular solution of the velocity function is:

$v(t) = t^{2} + 5\cdot t -5, \forall t \geq 0$

Position

$s(t) = \int\limits^{t}_{0} {v(t)} \, dt$

$s(t) = \int\limits^{t}_{0} {(t^{2}+5\cdot t -5)} \, dt$

$s(t) = \int\limits^{t}_0 {t^{2}} \, dt + 5\int\limits^{t}_0 {t} \, dt - 5\int\limits^{t}_0\, dt$

$s(t) = \frac{1}{3}\cdot t^{3} + \frac{5}{2}\cdot t^{2} - 5\cdot t + s(0)$

Where $s(0)$ is the initial position.

If $s(0) = 6$, the particular solution of the position function is:

$s(t) = \frac{1}{3}\cdot t^{3} + \frac{5}{2}\cdot t^{2} - 5\cdot t + 6,\forall t \geq 0$