Question

A point has simultaneous velocities represented by $u$ , $2u$ , $3\sqrt 3 u$ and $4u$ . The angles between the first and second, the second and third and the third and fourth are respectively ${60^ \circ }$ , ${90^ \circ }$ and \[{150^ \circ }\] . The angle the resultant velocity makes with $u$ is
A. ${120^ \circ }$
B. ${60^ \circ }$
C. ${150^ \circ }$
D. ${30^ \circ }$

Answer 1

Hint: We can find the angle between the resultant and the velocity $u$ by using the concept of resolution of the vectors along coordinate axes. Also, we use the concept of trigonometric ratios of simple angles and the resultant of two or more than two vectors.

Complete step by step answer:
Let us consider the point at X-axis and its velocity $u$ is along X-axis.
Resolving the components of $u$ along two axes, we get
$u = u\cos {0^ \circ }\hat i + u\sin {0^ \circ }\hat j = u\hat i - - - - - - (1)$
Given, the angle between $u$ and $2u$ is ${60^ \circ }$ . Components along $2u$.
$2u\cos {60^ \circ }\hat i + 2u\sin {60^ \circ }\hat j = u\hat i + u\sqrt 3 \hat j - - - - - - (2)$
Angle between $2u$ & $3\sqrt 3 u$ is ${90^ \circ }$. Therefore the angle between $u$ & $3\sqrt 3 u$ is $60 + 90 = {150^ \circ }$.
Resolving $3\sqrt 3 u$ , we get
$3\sqrt 3 u\cos {150^ \circ }\hat i + 3\sqrt 3 u\sin {60^ \circ }\hat j = - \dfrac{9}{2}u\hat i + \dfrac{{3\sqrt 3 }}{2}u\hat j - - - - - - (3)$

Again, the angle between $3\sqrt 3 u$ & $4u$ is ${150^ \circ }$ . Therefore angle between $u$ and $4u$ is $150 + 150 = {300^ \circ }$
Resolving $4u$, we get
$4u\cos {300^ \circ }\hat i + 4u\sin {300^ \circ }\hat j = 2u\hat i - 2\sqrt 3 u\hat j - - - - - - (4)$
Now, the resultant of all the velocities from eq $(1),(2),(3),(4)$ is given by
$R = u\hat i + u\hat i + u\sqrt 3 \hat j + - \dfrac{9}{2}u\hat i + \dfrac{{3\sqrt 3 }}{2}u\hat j + 2u\hat i - 2\sqrt 3 u\hat j$
$\therefore R = - \dfrac{1}{2}u\hat i + \dfrac{{\sqrt 3 }}{2}u\hat j - - - - - - - - (5)$
The angle between the resultant $R$ and velocity $u$ is given by
\[\theta = {\tan ^{ - 1}}\left\{ {\dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{{ - 1}}{2}}}} \right\}\]
\[\therefore \theta = {\tan ^{ - 1}}\left\{ { - \sqrt 3 } \right\} = {120^ \circ }\]

Hence, option A is correct.

Note: We must know all the trigonometric ratios of angles and their values.Resolving the vectors along two axes is essential for getting the result of the vector. The angle between the X-axis and a vector represented by resolution of that vector is equal to the $y$ component of the vector to the $x$ component of the vector along two axes.