Kinematics 2 Question 1
2. A particle is moving Eastwards with a velocity of $5 m / s$. In $10 s$, the velocity changes to $5 m / s$ Northwards. The average acceleration in this time is
(1982, 3M)
(a) zero
(b) $\frac{1}{\sqrt{2}} m / s^{2}$ towards North-Eeast
(c) $\frac{1}{\sqrt{2}} m / s^{2}$ towards North-West
(d) $\frac{1}{2} m / s^{2}$ towards North
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Answer:
Correct Answer: 2. (b)
Solution:
- Given, drag force, $F=m \psi^{2}$
As we know, general equation of force
$$ =m a $$
Comparing Eqs. (i) and (ii), we get
$$ a=\gamma v^{2} $$
$\therefore$ Net retardation of the ball when thrown vertically upward is
$$ \begin{aligned} a _{net} & =-\left(g+\gamma^{2}\right)=\frac{d v}{d t} \\ \Rightarrow \quad \frac{d v}{\left(g+\gamma^{2}\right)} & =-d t \end{aligned} $$
By integrating both sides of Eq. (iii) in known limits, i.e.
When the ball thrown upward with velocity $v _0$ and then reaches to its zenith, i.e. for maximum height at time $t=t, v=0$
$$ \begin{gathered} \Rightarrow \quad \int _{v _0}^{0} \frac{d v}{\left(Y v^{2}+g\right)}=\int _0^{t}-d t \\ \text { or } \quad \frac{1}{Y} \int _{v _0}^{0} \frac{1}{\sqrt{\frac{g}{Y}}{ }^{2}+v^{2}} d v=-\int _0^{t} d t \\ \Rightarrow \quad \frac{1}{Y} \cdot \frac{1}{\sqrt{g / Y}} \cdot \tan ^{-1} \frac{v}{\sqrt{g / Y}}=-t \\ \because \int _{v _0}^{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a} \end{gathered} $$
$$ \Rightarrow \quad \frac{1}{\sqrt{\mathrm{\gamma g}}} \cdot \tan ^{-1} \frac{\sqrt{Y} v _0}{\sqrt{g}}=t $$
