# Question #ab83f

Aug 24, 2016

Let

$\text{East"-> +"ve direction of x-axis}$

$\text{West"->-"ve direction of x-axis}$

$\text{North"-> +"ve direction of y-axis}$

$\text{South"->-"ve direction of y-axis}$

$\vec{{V}_{i}} \to \text{Initial velocity vetor}$

$\vec{{V}_{f}} \to \text{Final velocity vetor}$

$\left\mid \vec{{V}_{i}} \right\mid = 8.5 \frac{m}{s}$

$\left\mid \vec{{V}_{f}} \right\mid = 19 \frac{m}{s}$

$t \to \text{Time of change of velocity} = 2 s$

$\text{Direction of } \vec{{V}_{i}} = S {42}^{\circ} W = {\left(270 - 42\right)}^{\circ}$

$\text{Direction of } \vec{{V}_{f}} = W {35}^{\circ} N = {\left(180 - 35\right)}^{\circ}$

$\vec{{V}_{i}} = 8.5 \left(\cos \left(270 - 42\right) \hat{i} + \sin \left(270 - 42\right) \hat{j}\right)$
$= - 8.5 \left(\sin 42 \hat{i} + \cos 42 \hat{j}\right)$
$= - 5.69 \hat{i} - 6.32 \hat{j} \frac{m}{s}$

$\vec{{V}_{f}} = 19 \left(\cos \left(180 - 35\right) \hat{i} + \sin \left(180 - 35\right) \hat{j}\right)$
$= 19 \left(- \cos 35 \hat{i} + \sin 35 \hat{j}\right)$
$= - 15 , 56 \hat{i} + 10.9 \hat{j} \frac{m}{s}$

$\text{Now average acceleration}$
$\vec{a} = \frac{\vec{{V}_{f}} - \vec{{V}_{i}}}{t}$

$= \frac{1}{2} \left(- 9.87 \hat{i} + 17.22 \hat{j}\right) m {s}^{-} 2$

$\left\mid \vec{a} \right\mid = \frac{1}{2} \sqrt{{9.87}^{2} + {17.22}^{2}} \approx 19.85 m {s}^{-} 2$