A particle experiences an acceleration $\vec{a}=\alpha \vec{v}$, where $\vec{v}$ is the velocity of ...

Given, $\vec{a}=\alpha \vec{v}$$$ \begin{gathered} a=\alpha v \\ \frac{d v}{d t}=\alpha v \end{gathered} $$On Integrating both sides:$$ \begin{aligned} &\begin{gathered} \int \frac{d v}{v}=\int \alpha d t \\ \log _e(v)=\alpha t+c \end{gathered}\\ &\text { }\\ &\begin{gathered} v=e^{\alpha t+c}=e^{\alpha t} \cdot e^c \\ \frac{d S}{d t}=e^{\alpha t} \cdot e^c \end{gathered} \end{aligned} $$$$ \text { Integrate to find displacement: } $$$$ \begin{aligned} &\int d S=e^c \int e^{\alpha t} \cdot d t\\ &[S]_x^y=\left[\frac{e^c}{\alpha} \cdot e^{\alpha t}\right]_x^y \end{aligned} $$So,$$ \begin{gathered} S_{12}=[S]_{t_1}^{t_2}=\left[\frac{e^c}{\alpha} \cdot e^{\alpha t}\right]_{t_1}^{t_2} \\ S_{12}=\frac{e^c}{\alpha} \cdot\left(e^{\alpha t_2}-e^{\alpha t_1}\right) \\ S_{13}=\frac{e^c}{\alpha} \cdot\left(e^{\alpha t_3}-e^{\alpha t_1}\right) \end{gathered} $$$$ \begin{gathered} \frac{S_{13}}{S_{12}}=\frac{\frac{e^c}{\alpha} \cdot\left(e^{\alpha t_3}-e^{\alpha t_1}\right)}{\frac{e^c}{\alpha} \cdot\left(e^{\alpha t_2}-e^{\alpha t_1}\right)} \\ \frac{S_{13}}{S_{12}}=\frac{\left(e^{\alpha t_3}-e^{\alpha t_1}\right)}{\left(e^{\alpha t_2}-e^{\alpha t_1}\right)} \end{gathered} $$Taking $e^{\alpha t_1}$ common,$$ \begin{gathered} \frac{S_{13}}{S_{12}}=\frac{e^{t_1}\left(e^{\alpha t_3-t_1}-1\right)}{e^{t_1}\left(e^{\alpha t_2-t_1}-1\right)} \\ \frac{S_{13}}{S_{12}}=\frac{\left(e^{\alpha t_3-t_1}-1\right)}{\left(e^{\alpha t_2-t_1}-1\right)}=\frac{\exp \left(\alpha t_3-t_1\right)-1}{\exp \left(\alpha t_3-t_1\right)-1} \end{gathered} $$

A ball is thrown vertically upwards with an initial speed $u$ from a height $h$ above the ground. Th...

$$ \text { Let's break the motion down into two parts }(u \rightarrow 0,0 \rightarrow v) \text { and then evaluate using: } $$$$ s_{\text {avg }}=\frac{\text { total distance }}{\text { total time }} $$The total time in the ascent of the ball from the point (1) $\rightarrow$ (2) can be calculated by using $v=u+a t$ :$$ t_1=\frac{u}{g} $$similarly, (2)$$ \begin{aligned} & \rightarrow \text { (3) } t_2=\frac{v}{g} \\ & t_{\text {net }}=t_1+t_2=\frac{u}{g}+\frac{v}{g}=\frac{u+v}{g} \end{aligned} $$And the distance can be calculated using $v^2-u^2=2 a s$ $d_1=\frac{u^2}{2 g}$ with velocity $u$$d_2=\frac{v^2}{2 g}$ with velocity $v$$$ d_{\text {total }}=d_1+d_2=\frac{u^2}{2 g}+\frac{v^2}{2 g}=\frac{u^2+v^2}{2 g} $$Now, we can find the average speed by dividing the total distance by the total time:$$ \text { Average speed }=\frac{d_{\text {total }}}{t_{\text {total }}}=\frac{\frac{u^2+v^2}{2 g}}{\frac{u+v}{g}} $$Therefore, the average speed over the entire path is: $\frac{u^2+v^2}{2(u+v)}$

The velocity $v(t)$ of a particle moving in one dimension is given by:$$ v(t)= \begin{cases}\alpha t...

The velocity $ v(t) $ of a particle moving in one dimension is given as follows:$ v(t) = \begin{cases} \alpha t, & 0 \leq t \leq \frac{T}{3} \\ \frac{\alpha T}{3}, & \frac{T}{3} \leq t \leq \frac{2T}{3} \\ \alpha(T-t), & \frac{2T}{3} \leq t \leq T \end{cases} $Where $\alpha \neq 0$ is a constant. To find the total displacement of the particle from time $ t = 0 $ to $ t = T $, we use the integral of velocity to find displacement, $ s = \int v(t) \, dt $.Displacement Calculation:For $ 0 \leq t \leq \frac{T}{3} $:$ s_1 = \int_0^{\frac{T}{3}} \alpha t \, dt = \left[\frac{\alpha t^2}{2}\right]_0^{\frac{T}{3}} = \frac{\alpha}{2} \left(\frac{T^2}{9}\right) = \frac{\alpha T^2}{18} $For $ \frac{T}{3} \leq t \leq \frac{2T}{3} $:$ s_2 = \int_{\frac{T}{3}}^{\frac{2T}{3}} \frac{\alpha T}{3} \, dt = \left[\frac{\alpha T t}{3}\right]_{\frac{T}{3}}^{\frac{2T}{3}} = \frac{\alpha T}{9} $For $ \frac{2T}{3} \leq t \leq T $:$ s_3 = \int_{\frac{2T}{3}}^T \alpha(T-t) \, dt = \left[\alpha(Tt - \frac{t^2}{2})\right]_{\frac{2T}{3}}^T $Simplifying further:$ = \alpha \left[T \left(\frac{T}{3}\right) - \frac{1}{2}\left(T^2 - \frac{4T^2}{9}\right) \right] = \frac{\alpha T^2}{18} $Total Displacement:Adding all the displacements:$ s_{\text{net}} = s_1 + s_2 + s_3 = \frac{\alpha T^2}{18} + \frac{\alpha T}{9} + \frac{\alpha T^2}{18} = \frac{2 \alpha T^2}{9} $

Motion In A Straight Line - IAT-IISER Physics PYQs

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Motion In A Straight Line

High-Yield Trend

Questions
3 MCQs

Official Solution

Official Solution

Official Solution

Motion In A Straight Line

High-Yield Trend

Questions 3 MCQs

Official Solution

Official Solution

Official Solution

Questions
3 MCQs