# Variable acceleration?

## The velocity v ms-1 of a particle after t seconds is given by v=t−3t2 The initial displacement of the particle from the origin is 3m. The displacement of the particle from the origin after 5 seconds is ____m

Apr 19, 2018

$x = - \frac{219}{2} \text{ m}$

#### Explanation:

We know that $v = \frac{\mathrm{dx}}{\mathrm{dt}}$:

$\frac{\mathrm{dx}}{\mathrm{dt}} = t - 3 {t}^{2}$

where $x$ is the displacement

Integrating:

$x = \int t - 3 {t}^{2} \mathrm{dt}$

$x = {t}^{2} / 2 - {t}^{2} + {x}_{0}$

We are told that ${x}_{0} = 3 \text{ m}$

$x = {t}^{2} / 2 - {t}^{3} + 3 \text{ m}$

Evaluate at $t = 5 \text{ s}$

$x = {5}^{2} / 2 - {5}^{3} + 3 \text{ m}$

$x = - \frac{219}{2} \text{ m}$

Apr 19, 2018

the displacement of the particle after 5 seconds is
$s = - 109.5$

#### Explanation:

$v = t - 3 {t}^{2}$
$v = \frac{\mathrm{ds}}{\mathrm{dt}}$

$\frac{\mathrm{ds}}{\mathrm{dt}} = t - 3 {t}^{2}$

$\mathrm{ds} = \left(t - 3 {t}^{2}\right) \mathrm{dt}$
$\int \mathrm{ds} = \int \left(t - 3 {t}^{2}\right) \mathrm{dt}$

$s = {s}_{0} + \frac{1}{2} {t}^{2} - {t}^{3}$

$s = 3 , t = 0$

$3 = {s}_{0} + \frac{1}{2} {\left(0\right)}^{2} - {\left(0\right)}^{3}$

${s}_{0} = 3$

$s = 3 + \frac{1}{2} {t}^{2} - {t}^{3}$

Substitute t=5sec

$s = 3 + \frac{1}{2} {\left(5\right)}^{2} - {\left(5\right)}^{3}$

$s = 3 + \frac{25}{2} - 125$

$s = 3 + 12.5 - 125$

$s = 15.5 - 125$

$s = - 109.5$
Hence, the displacement of the particle after 5 seconds is
$s = - 109.5$