Motion in a Straight Line

Case 1: $h=-u{t}_1+\frac{1}{2}g{t}_1^2$ ( downward taken positive)

Case 2 $h=u{t}_2+\frac{1}{2}g{t}_2^2$

\begin{equation}
\begin{aligned}
& \text { } t_1=\frac{u+\sqrt{u^2+2 g h}}{g} \\
& t_2=\frac{-u+\sqrt{u^2+2 g h}}{g} \\
& \text{Case 3}\\ t=\frac{\sqrt{2 g h}}{g} \\
& t_1 t_2=\frac{\left(u^2+2 g h\right)-u^2}{g^2}=\frac{2 g h}{g^2}=t^2 \\
& \Rightarrow t=\sqrt{t_1 t_2}
\end{aligned}
\end{equation}

Sol. The time period of an orbiting body is related to its distance from the central body (Earth in this case) by Kepler’s third law, which states:

Where:

• $T$ = time period
• $R$ = distance from the central body

Given:

Let $T_{\text{moon}}$ and $T_{\text{satellite}}$​ be the time periods of the Moon and the satellite, respectively. Using Kepler's law:

Substitute values:

Simplify:

Correct answer: (a) 1 day

Sol. The time period of an orbiting body is related to its distance from the central body (Earth in this case) by Kepler’s third law, which states:

Where:

• $T$ = time period
• $R$ = distance from the central body

Given:

Let $T_{\text{moon}}$ and $T_{\text{satellite}}$​ be the time periods of the Moon and the satellite, respectively. Using Kepler's law:

Substitute values:

Simplify:

Correct answer: (a) 1 day

• Find initial velocity (v₀) and initial acceleration (a₀) from the position equation x(t).
• x = αt⁴ + βt² + γt + δ.
• Velocity v = dx/dt = 4αt³ + 2βt + γ. Initial velocity v₀ (at t=0) = γ.
• Acceleration a = dv/dt = 12αt² + 2β. Initial acceleration a₀ (at t=0) = 2β.
• Ratio v₀ : a₀ = γ : 2β.

$\overrightarrow{v} =\frac{d\overrightarrow{s}}{dt} =4t\hat{j}$

Hence, velocity of ant t = 1 sec

$V_{avg}=\frac{S}{{\displaystyle \frac{S}{2v_1}}+\left({\displaystyle \frac{S}{v_2+v_3}}\right)}$

$V_{avg}= \frac{2v_1\left(v_2+v_3\right)}{v_2+v_3+2v_1}$

$\overrightarrow{v} =\frac{d\overrightarrow{s}}{dt} =4t\hat{j}$

Hence, velocity of ant t = 1 sec

$V_{avg}=\frac{S}{{\displaystyle \frac{S}{2v_1}}+\left({\displaystyle \frac{S}{v_2+v_3}}\right)}$

$V_{avg}= \frac{2v_1\left(v_2+v_3\right)}{v_2+v_3+2v_1}$

\begin{equation}
\begin{aligned}
&\begin{aligned}
& v^2-u^2=2 a S \\
& \left(\frac{2 u}{3}\right)^2=u^2+2(-a)\left(4 \times 10^{-2}\right) \\
& \Rightarrow \frac{4 u^2}{9}=u^2-2 a\left(4 \times 10^{-2}\right) \\
& \Rightarrow-\frac{5 u^2}{9}=-2 a\left(4 \times 10^{-2}\right) \\
& \Rightarrow 0=\left(\frac{2 u}{3}\right)^2+2(-a)(x) \\
& \Rightarrow-\frac{4 u^2}{9}=-2 a x
\end{aligned}\\
&\text { Dividing equation (1) by (2), we get }\\
&\begin{aligned}
& \frac{5}{4}=\frac{4 \times 10^{-2}}{x} \Rightarrow x=\frac{16}{5} \times 10^{-2} \\
& \Rightarrow x=3.2 \times 10^{-2} \mathrm{~m}=32 \times 10^{-3} \mathrm{~m}
\end{aligned}
\end{aligned}
\end{equation}