# Question #05757

Feb 10, 2017

$6.8 {\text{ ms}}^{-} 1$

#### Explanation:

It is a case of non-uniform acceleration. (Assuming that $t$ is acceleration is time).
We know that
Acceleration $a = \frac{\mathrm{dv} \left(t\right)}{\mathrm{dt}} = - 0.9 t$
$\implies \mathrm{dv} \left(t\right) = - 0.9 t \cdot \mathrm{dt}$

Integrating both sides we get
$\implies v \left(t\right) = - 0.9 \int t \cdot \mathrm{dt}$
$\implies v \left(t\right) = - 0.9 \left({t}^{2} / 2 + C\right)$ .....(1)
where $C$ is a constant of integration. It is given that the initial velocity at $t = 0$ is $23 {\text{ ms}}^{-} 1$. Inserting in (1)
$v \left(0\right) = 23 = - 0.9 \left({0}^{2} / 2 + C\right)$
$\implies - 0.9 C = 23$
$\implies C = - \frac{23}{0.9}$
Expression for velocity becomes
$\implies v \left(t\right) = - 0.9 \left({t}^{2} / 2 - \frac{23}{0.9}\right)$
$\implies v \left(t\right) = - 0.45 {t}^{2} + 23$ ......(2)

To find the velocity after $6 s$. From (2) we get
$v \left(6\right) = - 0.45 \times {6}^{2} + 23$
$v \left(6\right) = 6.8 {\text{ ms}}^{-} 1$