Question

A ball is suspended by a thread from the ceiling of a car. The brakes are applied and the speed of the car changes uniformly from $\mathop {36kmh}\nolimits^{ - 1}$ to zero in $5s$ . The angle by which the ball deviates from the vertical ($g = 10m{s^{ - 2}}$) is:
(A) $ta{n^{ - 1}}(\frac{1}{3})$
(B) $si{n^{ - 1}}(\frac{1}{5})$
(C) $ta{n^{ - 1}}(\frac{1}{5})$
(D) $co{t^{ - 1}}(\frac{1}{3})$

Answer 1

Hint We will first find out the acceleration of the car. Finally we will find out the angle it makes with the vertical.
Formulae Used: $a = \frac{{(v - u)}}{t}$
Where, $a$ is the acceleration of the body, $v$ is the final velocity, $u$ is the initial velocity and $t$ is the time taken by the body.

Complete Step By Step Solution
Here,
$u = {\text{ }}36{\text{ }}km{h^{ - 1}} = 10m{s^{ - 1}}$
$v = 0km{h^{ - 1}} = 0m{s^{ - 1}}$
$t = 5s$
Now,
${a_{car}} = \frac{{(0 - 10)}}{5}$
Now,
${a_{}} = - 2m{s^{ - 2}}$

tan\theta = \frac{a}{g} = \frac{2}{{10}} = \frac{1}{5} \\\ \\\ \end{gathered} $$ Thus, the answer turns out to be $$\theta = ta{n^{ - 1}}(\frac{1}{5})$$ which is (C). But, Also, If we take $$\begin{gathered} sin\theta = \frac{2}{{10}} = \frac{1}{5} \\\ \\\ \end{gathered} $$ $$ \Rightarrow \theta = si{n^{ - 1}}(\frac{1}{5})$$ **Hence, (B) is also correct.** **Note** We got the value of $$sin\theta $$ , we took $${a_{net}} = \sqrt {{a^2} + {g^2}} $$ an d value turns out to be $$10m{s^{ - 2}}$$. Also, we took only the magnitude as it only plays the role to form the angle.