Motion in a Straight Line

Given the relationship between time $ t $ and position $ x $ of a moving particle:

$ t = x^2 + x $

First, find the derivative of $ t $ with respect to $ x $:

$ \frac{d t}{d x} = 2x + 1 $

The velocity $ v $ is the inverse of this derivative, as it's given by the derivative of position $ x $ with respect to time $ t $:

$ v = \frac{d x}{d t} = \frac{1}{2x + 1} $

Next, calculate the derivative of $ v $ with respect to $ x $:

$ \frac{d v}{d x} = \frac{-2}{(2x + 1)^2} $

The acceleration $ a $ is then the product of velocity $ v $ and the derivative of velocity with respect to $ x $:

$ a = v \cdot \frac{d v}{d x} = \frac{1}{2x + 1} \cdot \left(\frac{-2}{(2x + 1)^2}\right) $

Simplifying this expression, we find:

$ a = -\frac{2}{(2x + 1)^3} $

$\begin{aligned} & \text { Let velocity of bus }=v \mathrm{~km} / \mathrm{hr} \\ & \text { Relative velocity of bus w.r.t. scooty }=(v-60) \\ & \text { Distance between } 2 \text { consecutive buses }=v T \\ & (v-60) 30=v T \quad\text{.... (i)}\\ & Y \rightarrow X \\ & (v+60) 10=v T \quad\text{.... (ii)}\\ & \text { Equating }(1) \text { and }(2) \\ & (v-60) 30=(v+60) 10 \\ & \therefore \quad v=120 \mathrm{~km} / \mathrm{hr} \\ & T=15 \mathrm{~min} \end{aligned}$

To find the ratio of the initial velocity to the initial acceleration of a particle moving along the $x$-axis, we need to differentiate the given position function with respect to time $t$.

Let's start by writing the position function:

$x = \alpha t^4 + \beta t^2 + \gamma t + \delta$

To find the velocity ($v$), we differentiate $x$ with respect to $t$:

$v = \frac{dx}{dt} = \frac{d}{dt} (\alpha t^4 + \beta t^2 + \gamma t + \delta)$

This gives us:

$v = 4\alpha t^3 + 2\beta t + \gamma$

The initial velocity is the velocity at $t = 0$:

$v(0) = 4\alpha (0)^3 + 2\beta (0) + \gamma = \gamma$

Next, to find the acceleration ($a$), we differentiate $v$ with respect to $t$:

$a = \frac{dv}{dt} = \frac{d}{dt} (4\alpha t^3 + 2\beta t + \gamma)$

This gives us:

$a = 12\alpha t^2 + 2\beta$

The initial acceleration is the acceleration at $t = 0$:

$a(0) = 12\alpha (0)^2 + 2\beta = 2\beta$

Now, we find the ratio of the initial velocity to the initial acceleration:

$\text{Ratio} = \frac{v(0)}{a(0)} = \frac{\gamma}{2\beta}$

Therefore, the correct answer is:

Option D: $\gamma: 2\beta$

Initially, the body has zero velocity and zero slope. Hence the acceleration would be zero initially. After that, the slope of v-t curve is constant and positive.

After some time, velocity becomes constant and acceleration is zero.

After that, the slope of v-t curve is constant and negative.

$\begin{aligned} & \overrightarrow{\mathrm{V}}=\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+0 \hat{\mathrm{k}} \\ & \text { at } \mathrm{t}=1 \mathrm{sec} \\ & \overrightarrow{\mathrm{V}}=4 \hat{\mathrm{i}}+4(1) \hat{\mathrm{j}} \\ & |\overrightarrow{\mathrm{V}}|=\sqrt{4^2+4^2}=4 \sqrt{2} \\ & \tan \alpha=\frac{4}{4}=1 \\ & \alpha=45^{\circ}\end{aligned}$

$V_{\text {avg }}=\frac{2 v_{1} v_{2}}{v_{1}+v_{2}}=\frac{2(v)(2 v)}{v+2 v}=\frac{4 v^{2}}{3 v}=\frac{4 v}{3}$

Slope of position-time (x - t) graph represents velocity.

In graph-2, the slope is increasing hence the velocity is increasing hence acceleration is positive.

${S_{{n^{th}}}} = u + {1 \over 2}a(2n - 1)$

${S_{{1^{st}}}} = {1 \over 2}g(2 \times 1 - 1) = {g \over 2}$

${S_{{2^{nd}}}} = {1 \over 2}g(2 \times 2 - 1) = 3 \times \left( {{1 \over 2}g} \right)$

${S_{{3^{rd}}}} = {1 \over 2}g(2 \times 3 - 1) = 5 \times \left( {{1 \over 2}g} \right)$

${S_{{4^{th}}}} = {1 \over 2}g(2 \times 4 - 1) = 7 \times \left( {{1 \over 2}g} \right)$

${S_{{1^{st}}}}:{S_{{2^{nd}}}}:{S_{{3^{rd}}}}:{S_{{4^{th}}}}$

$ = 1:3:5:7$

Slope of x-t curves gives the velocity

$\Rightarrow$ Ratio $ = {{\tan 30^\circ } \over {\tan 45^\circ }} = {1 \over {{{\sqrt 3 } \over 1}}} = 1:\sqrt 3 $

When a particle is released from rest position under gravity, then $v=0$ but $a \neq 0$

Equation of given line when velocity is plotted against position with intercept $=v_0$ and slope $=v_0 / x_0$ is,

$v=-\frac{v_0 x}{x_0}+v_0$ ...... (i)

($-$Ve sign is became the slope is decreasing) On differentiating with respect to $t$

$\frac{d v}{d t}=-\frac{v_0}{x_0} \frac{d x}{d t}+0$

or $\quad a=-\frac{v_0}{x_0} v$ ...... (ii)(

Putting $v$ from Eq. (i) in Eq. (ii),

$\begin{aligned} a & =-\frac{v_0}{x_0}\left[-\frac{v_0}{x_0} x+v_0\right] \\ & =\frac{v_0^2}{x_0^2} x-\frac{v_0^2}{x_0} \end{aligned}$

Which represents a straight line with positive slope and negative intercept.

We have

$S=u t+\frac{1}{2} a t^2$

$\begin{aligned} \text{or} \quad h & =(0) t+\frac{1}{2} \times 9.8 \times(4)^2 \quad\left(\begin{array}{l} \because a=g \\ \because u=0 \end{array}\right) \\ & =78.4 \mathrm{~m} \end{aligned}$

$\begin{gathered} \text{Given,} \quad \Delta t=0.2 \mathrm{~s} \\ t=4 \mathrm{~s} \end{gathered}$

Now, for error

$\frac{\Delta h}{h}= \pm 2\left(\frac{\Delta t}{t}\right)= \pm 2\left(\frac{0.2}{4}\right)= \pm 0.1$

$\begin{aligned} \text{or} \quad \Delta h & = \pm 0.1 \times h \\ & = \pm 0.1 \times 78.4= \pm 7.84 \mathrm{~m} \end{aligned}$