Motion in a Straight Line

The juggler throws n balls per second.

$\therefore$ Interval between balls = ${1 \over n}$ seconds

At maximum height velocity of first ball $v = 0$

Juggler throws all balls with same initial velocity $ = u$

For 1st ball,

${u_1} = u$, ${v_1} = 0$, $a = - g$, $S = {H_{\max }}$

Using formula,

${v^2} = {u^2} + 2as$

$0 = {u^2} - 2g\,{H_{\max }}$

$ \Rightarrow {H_{\max }} = {{{u^2}} \over {2g}}$ ...... (1)

And using formula,

$v = u + at$

$ \Rightarrow 0 = u - gt$

$ \Rightarrow t = {u \over g}$

$\therefore$ Time taken by ball 1 to reach maximum height $({H_{\max }}) = {u \over g}$

According to the question,

$t = {1 \over n}$

$ \Rightarrow {u \over g} = {1 \over n}$

$ \Rightarrow u = {g \over n}$ ....... (2)

Putting value of $u$ in equation (1), we get

${H_{\max }} = {{{{\left( {{g \over n}} \right)}^2}} \over {2g}}$

$ = {{{g^2}} \over {2{n^2}g}}$

$ = {g \over {2{n^2}}}$

For first ${h \over 2}$ distance,

${h \over 2} = {1 \over 2}~gt_1^2$

$ \Rightarrow h = gt_1^2$

For total distance h,

$h = {1 \over 2}g{({t_1} + {t_2})^2}$

$ \Rightarrow gt_1^2 = {1 \over 2}g{({t_1} + {t_2})^2}$

$ \Rightarrow 2t_1^2 = {({t_1} + {t_2})^2}$

$ \Rightarrow \sqrt 2 {t_1} = {t_1} + {t_2}$

$ \Rightarrow (\sqrt 2 - 1){t_1} = {t_2}$

A ball is thrown vertically upward with a certain velocity, reaching a maximum height $ h $. We need to find the ratio of the times it is at height $ \frac{h}{3} $ while ascending and descending, respectively.

The initial velocity of the ball $ v $ can be given by:

$ v = \sqrt{2gh} $

When the ball is at height $ \frac{h}{3} $, the equation of motion is:

$ \frac{h}{3} = \sqrt{2gh} \cdot t - \frac{1}{2}gt^2 $

Rearranging the equation, we get a quadratic equation in terms of $ t $:

$ \frac{g}{2} t^2 - \sqrt{2gh} \cdot t + \frac{h}{3} = 0 $

The ratio of the times taken to ascend and descend to the height $ \frac{h}{3} $ can be found using the quadratic formula solution for $ t $, considering the corresponding velocities:

$ \frac{t_1}{t_2} = \frac{\sqrt{2gh} + \sqrt{2gh - \frac{2gh}{3}} }{\sqrt{2gh} - \sqrt{2gh - \frac{2gh}{3}}} $

Simplifying the terms inside the fraction:

$ = \frac{\sqrt{2} + \frac{2}{\sqrt{3}}}{\sqrt{2} - \frac{2}{\sqrt{3}} } = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} $

A NCC parade is moving at a steady speed of 9 km/h under a mango tree where a monkey is perched at a height of 19.6 meters. Suddenly, the monkey drops a mango. To determine which cadet will catch the mango, we need to calculate the distance of the cadet from the tree at the moment the mango is dropped, considering the acceleration due to gravity $ g = 9.8 \, \text{m/s}^2 $.

First, we use the formula to find the time $ t $ it takes for the mango to fall:

$ H = \frac{1}{2}\,g\,t^2 $

Substituting the given values:

$ 19.6 = 4.9\,t^2 $

Solving for $ t $:

$ t^2 = \frac{19.6}{4.9} $

$ t^2 = 4 $

$ t = 2 \, \text{sec} $

Next, we need to find the distance $ D $ the cadet travels in these 2 seconds:

$ D = \text{speed} \times \text{time} $

Since the speed is given in km/h, we first convert it to m/s:

$ 9 \, \text{km/h} = 9 \times \frac{1000}{3600} \, \text{m/s} = 2.5 \, \text{m/s} $

So, the distance covered by the cadet is:

$ D = 2.5 \, \text{m/s} \times 2 \, \text{sec} = 5 \, \text{m} $

Thus, the cadet who is 5 meters away from the tree at the moment the mango is dropped will catch the mango.

S = 4 cm

$v{'_4} = {v \over 3}$, a = constant

${v_{4 + x}} = 0$

$\left( {{v^2} - {{{v^2}} \over a}} \right) = 2a(4)$

$({v^2} - 0) = 2a(4 + x)$

${4 \over {4 + x}} = {8 \over 9}$

$ \Rightarrow x = 0.5$ m

$|{v_{10}}| = (100 + 10 \times 10)$ m/s

${v_{10}} = - 200$ m/s and ${v_0} = - 100$ m/s

from 10s to 20s velocity remains zero

$\Rightarrow$ from t = 0s to 10s velocity increases in magnitude linearly.

$\Rightarrow$ graph given in option A fits correctly.

A small toy begins to move from a standstill with a constant acceleration. It covers a distance of 10 meters in t seconds. We need to determine the distance the toy will travel in the subsequent t seconds.

First, we know that the initial distance traveled is given by the formula for constant acceleration starting from rest:

$\frac{1}{2} a t^2 = 10 \text{ m}$

Next, we calculate the total distance traveled after 2t seconds:

$\frac{1}{2} a (2t)^2 = \frac{1}{2} a \cdot 4t^2 = 2 a t^2$

Since we know from the initial condition that $ \frac{1}{2} a t^2 = 10 \text{ m} $, multiplying it by 4 gives

$2 a t^2 = 40 \text{ m}$

Thus, the additional distance traveled in the next t seconds is:

$\text{Total distance after 2t seconds} - \text{Distance already traveled in t seconds}$

$= 40 \text{ m} - 10 \text{ m}$

$= 30 \text{ m}$

${X_P} = \alpha t + \beta {t^2}$

${X_Q} = ft - {t^2}$

$\therefore$ ${V_P} = \alpha + 2\beta t$

${V_Q} = f - 2t$

$\because$ ${V_P} = {V_Q}$

$ \Rightarrow \alpha + 2\beta t = f - 2t$

$ \Rightarrow t = {{f - \alpha } \over {2(1 + \beta )}}$

The relationship between time $ t $ and distance $ x $ for a moving body is given by $ t = mx^2 + nx $, where $ m $ and $ n $ are constants. To determine the retardation (negative acceleration) of the motion, let's follow the steps to derive it:

Given:

$ t = mx^2 + nx $

First, differentiate $ t $ with respect to $ x $:

$ \frac{dt}{dx} = 2mx + n $

Since velocity $ v $ is defined as $ \frac{dx}{dt} $, we can write:

$ \frac{1}{v} = \frac{dt}{dx} = 2mx + n $

Thus,

$ v = \frac{1}{2mx + n} $

Next, to find the acceleration $ a $ (which is the derivative of velocity with respect to time), we start with the chain rule:

$ \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} $

Since $ v = \frac{dx}{dt} $, substituting $ x $ gives:

$ \frac{dx}{dt} = v $

Rewrite it:

$ \frac{dv}{dt} = \frac{dv}{dx} \cdot v $

Differentiate $ v $ with respect to $ x $:

$ v = (2mx + n)^{-1} $

$ \frac{dv}{dx} = -\frac{2m}{(2mx + n)^2} $

Then,

$ \frac{dv}{dt} = -\frac{2m}{(2mx + n)^2} \cdot v $

Substitute $ v = \frac{1}{2mx + n} $ into the expression:

$ \frac{dv}{dt} = -\frac{2m}{(2mx + n)^2} \cdot \frac{1}{2mx + n} $

Simplify the expression:

$ \frac{dv}{dt} = -2m \left( \frac{1}{2mx + n} \right)^3 $

$ a = -2m v^3 $

So, the retardation (negative acceleration) is:

$ a = -2m v^3 $

The slope of the given v versus x graph is $m = - {{{v_0}} \over {{x_0}}}$ and intercept is c = + v₀. Hence, v varies with x as

$v = - \left( {{{{v_0}} \over {{x_0}}}} \right)x + {v_0}$ ...... (1)

where v₀ and x₀ are constants of motion. Differentiating with respect to time t, we have

${{dv} \over {dt}} = - \left( {{{{v_0}} \over {{x_0}}}} \right){{dx} \over {dt}}$

or $a = - \left( {{{{v_0}} \over {{x_0}}}} \right)v$ ........ (2)

Using Eq. (1) in Eq. (2), we get

$a = - \left( {{{{v_0}} \over {{x_0}}}} \right)\left( { - {{{v_0}} \over {{x_0}}}x + {v_0}} \right)$

$a = {\left( {{{{v_0}} \over {{x_0}}}} \right)^2}x - {{v_0^2} \over {{x_0}}}$

Thus the graph of a versus x is a straight line having a positive slope = ${\left( {{{{v_0}} \over {{x_0}}}} \right)^2}$ and negative intercept = $ - {{v_0^2} \over {{x_0}}}$. Hence the correct choice is (a).

The velocity of a particle is given by $ v = v_0 + gt + Ft^2 $. Its position is $ x = 0 $ at $ t = 0 $. To find its displacement after time $ t = 1 $, follow these steps:

Given:

$ v = v_0 + gt + Ft^2 $

We know that:

$ \frac{dx}{dt} = v_0 + gt + Ft^2 $

To find the displacement, integrate both sides with respect to $ t $:

$ \int_{x = 0}^{x} dx = \int_{t = 0}^{t = 1} (v_0 + gt + Ft^2) \, dt $

This simplifies to:

$ x = \left[ v_0 t + \frac{gt^2}{2} + \frac{Ft^3}{3} \right]_{t = 0}^{t = 1} $

Evaluating the integral from $ t = 0 $ to $ t = 1 $:

$ x = v_0 + \frac{g}{2} + \frac{F}{3} $

Therefore, the displacement after time $ t = 1 $ is:

$ x = v_0 + \frac{g}{2} + \frac{F}{3} $