# Question #bb9d5

Mar 1, 2017

(a) From Newton's Second law of motion we know that
$\text{Force"="mass"xx"acceleration}$
Solving for acceleration Inserting given values we get
$\vec{a} = \frac{\vec{F}}{m}$
$\implies \vec{a} = \frac{1}{10} \left(9 \hat{i} + 12 \hat{j}\right)$
$\implies | \vec{a} | = \sqrt{{\left[\frac{9}{10}\right]}^{2} + {\left[\frac{12}{10}\right]}^{2}}$
$\implies | \vec{a} | = \frac{1}{10} \sqrt{81 + 144}$
$\implies | \vec{a} | = \frac{1}{10} \sqrt{225}$
$\implies | \vec{a} | = 1.5 m {s}^{-} 2$

(b) (i) Using the kinematic expression
$\vec{s} \left(t\right) = \vec{{s}_{0}} + \vec{u} t + \frac{1}{2} \vec{a} {t}^{2}$
$\vec{s} \left(5\right) = 0 + \left(\frac{11}{5} \hat{i} + \hat{j}\right) \times 5 + \frac{1}{2} \times \frac{1}{10} \left(9 \hat{i} + 12 \hat{j}\right) \times {5}^{2}$
$\implies \vec{s} \left(5\right) = 11 \hat{i} + 5 \hat{j} + \frac{45}{4} \hat{i} + 15 \hat{j}$
$\implies \vec{s} \left(5\right) = \left(11 + \frac{45}{4}\right) \hat{i} + 20 \hat{j}$
$\implies \vec{s} \left(5\right) = \frac{89}{4} \hat{i} + 20 \hat{j}$
$\implies | \vec{s} \left(5\right) | = \sqrt{{\left(\frac{89}{4}\right)}^{2} + {\left(20\right)}^{2}}$
$\implies | \vec{s} \left(5\right) | = \sqrt{\frac{7921}{16} + 400}$
$\implies | \vec{s} \left(5\right) | = \frac{\sqrt{14321}}{4} m$

(ii) Using the kinematic equation
$\vec{v} \left(t\right) = \vec{u} + \vec{a} t$
Inserting given values we get
$\vec{v} \left(t\right) = \left(\frac{11}{5} \hat{i} + \hat{j}\right) + \frac{1}{10} \left(9 \hat{i} + 12 \hat{j}\right) t$

(iii) North east direction is defined by a unit vector as
$\left(\hat{i} + \hat{j}\right)$
General expression for velocity is
$\vec{v} \left(t\right) = \left(\frac{11}{5} \hat{i} + \hat{j}\right) + \frac{1}{10} \left(9 \hat{i} + 12 \hat{j}\right) t$
To meet the condition we get at time $t$
$\vec{v} \left(t\right) = n \left(\hat{i} + \hat{j}\right)$
where $n$ is a positive number. Comparing two expressions for $\vec{v} \left(t\right)$ we get

$\frac{11}{5} + \frac{9}{10} t = n$
$\implies 10 n - 9 t = 22$ ......(1)
also
$1 + \frac{6}{5} t = n$
$\implies 5 n - 6 t = 5$ ......(2)
To solve (1) and (2), multiply (2) with $2$ and subtract from (1)
$\implies 10 n - 12 t = 10$ .....(3)
$3 t = 22 - 10 = 12$
$t = 4 s$