Motion in a Straight Line

The original position at $\mathrm{t}=0$ is $\mathrm{x}(0)=4(0)^3-3(0)=0$.

To find when it returns to the origin :

$ 4 t^3-3 t=0 $

$\Rightarrow $ $t\left(4 t^2-3\right)=0$

$ 4 t^2=3 \Rightarrow t^2=0.75 \Rightarrow t=\sqrt{0.75} \approx 0.866 \mathrm{~s} $

Hence, the Statement A is correct.

A turning point occurs when velocity $\mathrm{v}(\mathrm{t})=0$.

$ v(t)=\frac{d x}{d t}=12 t^2-3 $

At turning point,

$ \mathrm{v}(\mathrm{t})=0 \Rightarrow 12 \mathrm{t}^2=3 \Rightarrow \mathrm{t}^2=0.25 \Rightarrow \mathrm{t}=0.5 \mathrm{~s} $

So, the position at turning point is,

$ x(0.5)=4(0.5)^3-3(0.5)=4(0.125)-1.5=0.5-1.5=-1.0 \mathrm{~m} $

The distance from the origin is $|-1.0|=1$ unit.

Hence, the statements B is correct and statement D is incorrect. Statement E is incorrect because the particle does turn back at $\mathrm{t}=0.5 \mathrm{~s}$.

Acceleration $\mathrm{a}(\mathrm{t})$ is the derivative of velocity:

$ a(t)=\frac{d v}{d t}=24 t $

Since the particle starts moving from $\mathrm{t}=0$, for all $\mathrm{t} \geq 0$, the value of 24 t will always be greater than or equal to zero.

Hence, statement C is correct.

Therefore, statements $\mathrm{A}, \mathrm{B}$, and C and D are correct. The option 2 best represents the correct set.

Hence, the correct option is (B).

The first drop falls a distance of $H=5 \mathrm{~m}$ starting from rest $(u=0)$.

Using the equation of motion:

$ \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^2 $

$\Rightarrow $ $5=0+\frac{1}{2} \times 10 \times \mathrm{T}^2$

$\Rightarrow $ $5=5 \mathrm{~T}^2 \Rightarrow \mathrm{~T}^2=1 \Rightarrow \mathrm{~T}=1 \mathrm{~s}$

When the first drop hits the ground, the sixth drop is just beginning to fall. This means there are 5 equal time intervals between the first and sixth drops.

$ \Delta \mathrm{t}=\frac{\mathrm{T}}{5}=\frac{1}{5}=0.2 \mathrm{~s} $

At the instant the first drop strikes the ground $(\mathrm{t}=1 \mathrm{~s})$ :

Drop 1 has fallen for 1.0 s

Drop 2 has fallen for 0.8 s

Drop 3 has fallen for 0.6 s

Drop 4 has fallen for 0.4 s

Using the distance formula for the fourth drop:

$ \mathrm{h}_4=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_4\right)^2 $

$\Rightarrow $ $h_4=\frac{1}{2} \times 10 \times(0.4)^2$

$\Rightarrow $ $ \mathrm{h}_4=5 \times 0.16=0.8 \mathrm{~m} $

The height from the ground is the total height minus the distance fallen:

Height from ground $=\mathrm{H}-\mathrm{h}_4=5 \mathrm{~m}-0.8 \mathrm{~m}=4.2 \mathrm{~m}$

Hence, the correct option is (C).

For the given graph shows a straight line with a negative slope and a positive y -intercept.

We can represent this line using the standard equation for a straight line, $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ :

$ \mathrm{v}=-\mathrm{mx}+\mathrm{v}_0 $

Where:

• 1. $-m$ is the constant negative slope $(m>0)$.

• 2. $\mathrm{v}_0$ is the initial velocity at $\mathrm{x}=0$ (the y -intercept).

Acceleration a can be expressed in terms of velocity v and position x using the following formula :

$ a=v \frac{d v}{d x} $

$\Rightarrow $ $\mathrm{a}=\left(-\mathrm{mx}+\mathrm{v}_0\right) \frac{\mathrm{d}}{\mathrm{dx}}\left[-\mathrm{mx}+\mathrm{v}_0\right]$

$\Rightarrow $ $\mathrm{a}=\left(-\mathrm{mx}+\mathrm{v}_0\right)[-\mathrm{m}]$

$\Rightarrow $ $ \mathrm{a}=\mathrm{m}^2 \mathrm{x}-\mathrm{mv}_0 $

The equation $\mathrm{a}=\mathrm{m}^2 \mathrm{x}-\mathrm{mv}_0$ (linear function of x ) is also the equation of a straight line ( $\mathrm{y}=\mathrm{MX}+\mathrm{C}$ ) where,

• Slope, $\mathrm{M}=\mathrm{m}^2>0$. So the slope is positive.

• Intercept, $\mathrm{C}=-\mathrm{mv}_0$. Since both m and $\mathrm{v}_0$ are positive values from our initial graph, the intercept is negative.

So, the resulting graph is,

Therefore, the correct graph representing the acceleration versus distance variation is option D.

Hence, the correct option is (D).

Interval 1 : Free Fall (Before opening the parachute)

The paratrooper is falling under gravity with $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ starting from rest ( $\mathrm{u}_1=0$ ).

$ \text { Time }=\mathrm{t}_1=2 \mathrm{~s} $

Using equation of motion, the distance covered is

$ \mathrm{h}_1=\mathrm{u}_1 \mathrm{t}_1+\frac{1}{2} \mathrm{gt}_1^2 $

$\Rightarrow $$ \mathrm{h}_1=0+\frac{1}{2}(10)(2)^2=20 \mathrm{~m} $

Let the velocity at the end of interval 1 is $v_1$

$ \mathrm{v}_1=\mathrm{u}_1+\mathrm{gt}_1=0+(10)(2)=20 \mathrm{~m} / \mathrm{s} $

Interval 2 : Deacceleration (After opening the parachute)

The parachute opens, and the paratrooper starts deaccelerating until reaching a height of 10 m above the ground.

Initial velocity for this interval $=\mathrm{u}_2=20 \mathrm{~m} / \mathrm{s}$ (velocity $\mathrm{v}_1$ from interval 1)

Final velocity $=\mathrm{v}_2=5 \mathrm{~m} / \mathrm{s}$

Acceleration $\mathrm{a}=-3 \mathrm{~m} / \mathrm{s}^2$ (deacceleration)

So, the distance covered in interval 2, using the third equation of motion :

$ \mathrm{v}_2^2=\mathrm{u}_2^2+2 \mathrm{ah}_2 $

$\Rightarrow $ $(5)^2=(20)^2+2(-3) \mathrm{h}_2$

$\Rightarrow $ $25=400-6 \mathrm{~h}_2$

$\Rightarrow $ $ \mathrm{h}_2=\frac{375}{6}=62.5 \mathrm{~m} $

The total initial height,

$ \mathrm{H}=\mathrm{h}_1+\mathrm{h}_2+\mathrm{h}_{\text {ground }} $

$\Rightarrow $ $\mathrm{H}=20+62.5+10$

$\Rightarrow $ $ \mathrm{H}=92.5 \mathrm{~m} $

So, the airplane was at a height of 92.5 m when the paratrooper jumped.

Hence, the correct option is (A).

When an object is dropped from a moving carrier (like the balloon), it gets the velocity of that carrier at the exact moment of release.

Initial velocity of the stone (balloon) is $\mathrm{u}_{\mathrm{s}}=10 \mathrm{~m} / \mathrm{s}$ (upwards).

Initial height of the stone is $\mathrm{h}=75 \mathrm{~m}$.

The acceleration due to gravity is $\mathrm{g}=-10 \mathrm{~m} / \mathrm{s}^2$ (downwards).

Using the second equation of motion for the stone's displacement along vertical direction,

$ \mathrm{y}=\mathrm{u}_{\mathrm{y}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^2 $

Since the stone eventually hits the ground (at height 0 ), its displacement from the starting point ( 75 m ) is $\mathrm{y}=$ -75 m .

$ -75=(10) \mathrm{t}+\frac{1}{2}(-10) \mathrm{t}^2 $

$\Rightarrow $ $-75=10 t-5 t^2$

$\Rightarrow $ $15=-2 t+t^2$

$\Rightarrow $ $t^2-2 t-15=0$

$\Rightarrow $ $ (t-5)(t+3)=0 $

Since time cannot be negative, we take $\mathrm{t}=5$ seconds.

While the stone is falling for 5 seconds, the balloon continues to move upward at a constant velocity.

The constant velocity of balloon is $\mathrm{v}_{\mathrm{b}}=10 \mathrm{~m} / \mathrm{s}$.

The time interval of journey is $\mathrm{t}=5$ seconds.

Displacement $=$ Velocity × Time

Displacement $=10 \times 5=50 \mathrm{~m}$

The final height is the initial height plus the displacement covered during the 5 seconds.

Total Height $=75 \mathrm{~m}+50 \mathrm{~m}=125 \mathrm{~m}$

Therefore, the height of the balloon when the stone hits the ground is 125 m .

Hence, the correct option is (D).

Take the direction of motion of the cars as positive.

Speeds of the cars are:

$ v_A = 100 \text{ km/h}, \qquad v_B = 80 \text{ km/h} $

A person in car $B$ throws a stone towards car $A$.

Let the speed of throw relative to car $B$ be $v$ km/h.

So, speed of the stone relative to ground will be:

$ v_s = 80 + v \text{ km/h} $

Now, the stone hits car $A$ with a speed of $5 \text{ m/s}$ relative to car $A$.

Convert $5 \text{ m/s}$ into km/h:

$ 5 \times \frac{18}{5} = 18 \text{ km/h} $

Since the stone must catch car $A$ from behind, its speed relative to car $A$ is:

$ v_s - v_A = 18 $

Substitute values:

$ (80 + v) - 100 = 18 $

$ v - 20 = 18 $

$ v = 38 $

Hence, the required value of $v$ is

$ \boxed{38 \text{ km/h}} $

So, the correct option is Option C.

Average Velocity $\langle \vec{v} \rangle$: Calculated using the formula

$ \langle \vec{v} \rangle = \frac{\Delta \overrightarrow{s}}{\Delta t} = \frac{S_f - S_i}{t_f - t_i} $

where $S_f$ and $S_i$ are the final and initial displacements, respectively, and $t_f$ and $t_i$ are the final and initial times.

Instantaneous Velocity $\overrightarrow{v}$: Given by the slope of the displacement vs. time graph at any specific time:

$ \overrightarrow{v} = \frac{ds}{dt} = \text{slope at that point} $

Analysis of Each Time Interval:

(A) From $0$ to $3 \, \text{sec}$:

$ \langle \vec{v} \rangle = \frac{5 - 0}{3} = \frac{5}{3} \, \text{m/s} $

The statement that the average velocity is $10 \, \text{m/s}$ is incorrect based on our calculation.

(B) From $3$ to $5 \, \text{sec}$:

$ \langle \vec{v} \rangle = \frac{5 - 5}{2} = 0 \, \text{m/s} $

The average velocity is indeed $0 \, \text{m/s}$.

(C) At $t = 2 \, \text{sec}$:

The slope of the tangent line at $t = 2$ indicates that $\overrightarrow{v} = 5 \, \text{m/s}$. Therefore, this statement is correct.

(D) From $5$ to $7 \, \text{sec}$:

$ \langle \vec{v} \rangle = \frac{0 - 5}{2} = -2.5 \, \text{m/s} $

At $t = 6.5 \, \text{sec}$, the graph's slope shows that $\overrightarrow{v} = 10 \, \text{m/s}$, hence the statement is incorrect as the values are not equal.

(E) From $t = 0$ to $t = 9 \, \text{sec}$:

$ \langle \vec{v} \rangle = \frac{0 - 0}{9} = 0 \, \text{m/s} $

This statement is verified correct as the average velocity is zero over the entire period.

Based on the analysis, statements (B), (C), and (E) are correct.

To determine the acceleration of the particle, we start by differentiating the displacement function with respect to time $ t $.

The displacement of the particle is given by:

$ x = c_0(t^2 - 2) + c(t - 2)^2 $

First, compute the velocity $ v $ by differentiating $ x $ with respect to $ t $:

$ v = \frac{dx}{dt} = \frac{d}{dt}[c_0(t^2 - 2) + c(t - 2)^2] $

This gives:

$ v = 2c_0 t + 2c(t - 2) $

Next, determine the acceleration $ a $ by differentiating the velocity function with respect to time $ t $:

$ a = \frac{dv}{dt} = \frac{d}{dt}[2c_0 t + 2c(t - 2)] $

Calculating this gives:

$ a = 2c_0 + 2c $

Hence, the acceleration of the particle is $ 2c_0 + 2c $.

For option (A)

$\phi=\mathrm{kt}+\mathrm{C}$ it can be 1 D motion

$\mathrm{eg} \rightarrow \mathrm{x}=\mathrm{A} \sin \phi(\mathrm{SHM})$

For option (B)

$\mathrm{v}^2+\mathrm{x}^2=$ constant yes 1D

For option (C)

time can't be negative Not possible

For option (D)

Possible

A, B & D only

$\begin{aligned} & \text { Distance }=\text { Area under } \mathrm{v} \mathrm{v} / \mathrm{s} \text { t graph } \\ & \text { Distance }=\frac{1}{2} \times 2 \times 10+2 \times 10=30 \mathrm{~m}\end{aligned}$

The distance covered by the airplane in the first 30.5 seconds can be determined by calculating the area under the velocity-time graph.

The formula for distance in this scenario is:

$ \text{Distance} = \text{Area under the graph} $

Calculating the individual areas:

$ 300 \, \text{m/s} \times 2 \, \text{s} = 600 \, \text{m} $

$ 400 \, \text{m/s} \times 28.5 \, \text{s} = 11400 \, \text{m} $

Adding these values gives the total distance:

$ \text{Total distance} = 600 \, \text{m} + 11400 \, \text{m} = 12000 \, \text{m} $

Thus, the distance covered by the airplane in the first 30.5 seconds is $ 12000 \, \text{m} $, which is equivalent to 12 km.

$ \begin{aligned} & \text { } \quad S_n=u+\frac{1}{2} a(2 n-1) \\ & \therefore \quad S_3=0+\frac{1}{2} a(2 \times 3-1) \\ & S_3=\frac{5 a}{2} \text { and } S_5=0+\frac{1}{2} a(2 \times 5-1) \\ & S_5=\frac{9 a}{2} \\ & \text { but } S_3=10 \mathrm{~m} \text { and } S_5=18 \mathrm{~m} \\ & \therefore \frac{5 a}{2}=10 \Rightarrow a=4 \mathrm{~m} / \mathrm{s}^2 \end{aligned} $

Speed of the particle at time $t=4 \mathrm{~s}$,

$ v=u+a t=0+4 \times 4=16 \mathrm{~m} / \mathrm{s} $

$ \begin{aligned} & \text { } t=2 x^2+3 x \\ & \qquad \begin{aligned} & 1=4 x \frac{d x}{d t}+3 \frac{d x}{d t} \\ & v=\frac{d x}{d t}=\frac{1}{4 x+3} \\ & a=\frac{d v}{d t}=\frac{d}{d t}\left[\frac{1}{4 x+3}\right] \\ & a=-\frac{4}{(4 x+3)^2} \cdot \frac{d x}{d t}=-\frac{4}{(4 x+3)^3} \\ & \text { at } x=25 \mathrm{~cm}=0.25 \mathrm{~m} \\ & a=-\frac{4}{(4)^3}=-\frac{1}{16} \mathrm{~ms}^{-2} \end{aligned} \end{aligned} $

Step 1: Calculate the speed after falling 20 m

The person starts from rest, so the initial speed is 0. We use the formula:

$ \begin{aligned} v^2 & =u^2+2 g h \\ v & = \sqrt{2 \times 10 \times 20} \\ v & = \sqrt{400} = 20\ \mathrm{m/s} \end{aligned} $

This means after falling 20 meters, the person is moving at 20 m/s.

Step 2: Find the time to fall the first 20 m

Use the formula for distance with constant acceleration from rest:

$ \begin{aligned} s & = u t + \frac{1}{2}gt^2 \\ 20 & = 0 + \frac{1}{2} \cdot 10 \cdot t^2 \\ 20 & = 5 t^2 \\ t^2 & = 4 \\ t & = 2\ \text{seconds} \end{aligned} $

So, it takes 2 seconds to fall the first 20 meters.

Step 3: Calculate the distance left after 20 m

The total height is 2000 meters. After falling 20 meters, the remaining distance is:

$ \begin{aligned} \text{Remaining distance} &= 2000 - 20 \\ &= 1980\,\mathrm{m} \end{aligned} $

Step 4: Find the time to travel the remaining 1980 m at constant speed

Now, the person moves at 20 m/s (from step 1). Time taken is:

$ t = \frac{1980}{20} = 99\ \text{seconds} $

Step 5: Add both times to get the total time to reach the ground

Total time $= 2 + 99 = 101$ seconds.

Distance covered by the particle in first 3 seconds.

$ \begin{aligned} s & =u t+\frac{1}{2} a t^2 \\ & =0 \times 3+\frac{1}{2} \times 2 \times 3^2 \\ & =9 \mathrm{~m} \end{aligned} $

Velocity of particle after 3 s .

$ v=u+a t=0+2 \times 3=6 \mathrm{~m} / \mathrm{s} $

After $t=3 \mathrm{~s}$, direction of acceleration is reversed, so new acceleration,

$ a^{\prime}=-2 \mathrm{~m} / \mathrm{s}^2 $

Since, particle returns to its initial position, thus,

$ \begin{aligned} & s_1+s_2=0 \\ & s_2=-s_1=-s=-9 \mathrm{~m} \\ \therefore \quad & s_2=v t^{\prime}+\frac{1}{2} a^{\prime}\left(t^{\prime}\right)^2 \end{aligned} $

$ \begin{aligned} & -9=6 t^{\prime}+\frac{1}{2} \times(-2)\left(t^{\prime}\right)^2 \\ \Rightarrow \quad & \left(t^{\prime}\right)^2-6 t^{\prime}-9=0 \\ \Rightarrow \quad & t^{\prime}=\frac{-(-6) \pm \sqrt{36+36}}{2 \times 1} \end{aligned} $

$ \begin{aligned} & =3 \pm 3 \sqrt{2} \\ \Rightarrow \quad t^{\prime} & =(3+3 \sqrt{2}) \mathrm{s} \end{aligned} $

The total time from the beginning of the motion until the particle returns to its initial position,

$ \begin{aligned} & T=3+t^{\prime} \\ & =3+3+3 \sqrt{2}=6+3 \sqrt{2} \\ & =3(2+\sqrt{2}) \mathrm{s} \end{aligned} $

Time taken $\left(t_1\right)$ by freely falling body to travel first 5 m is given as,

$ \begin{aligned} s & =u t_1+\frac{1}{2} g t_1^2 \\ 5 & =0 \times t_1+\frac{1}{2} \times 10 \times t_1^2 \\ \Rightarrow \quad t_1 & =1 \mathrm{~s} \end{aligned} $

Time taken ( $t^{\prime}$ ) to travel 10 m .

$ \begin{aligned} 10 & =0 \times t_1^{\prime}+\frac{1}{2} \times 10 \times t^{\prime 2} \\ \Rightarrow \quad t^{\prime} & =\sqrt{2} \mathrm{~s} \end{aligned} $

$\therefore$ Time taken $\left(t_2\right)$ to travel next 5 m ,

$ t_2=t^{\prime}-t_1=(\sqrt{2}-1) \mathrm{s} $

Time taken $\left(t^{\prime \prime}\right)$ to travel 15 m ,

$ \begin{aligned} 15 & =0 \times t^{\prime \prime}+\frac{1}{2} \times 16 \times\left(t^{\prime \prime}\right)^2 \\ 15 & =5\left(t^{\prime \prime}\right)^2 \\ \therefore \quad t^{\prime \prime} & =\sqrt{3} \mathrm{~s} \end{aligned} $

Thus, time taken $\left(t_3\right)$ to travel lost 5 m

$ \begin{aligned} t_3 & =t^{\prime \prime}-t^{\prime}=\sqrt{3}-\sqrt{2} \\ \therefore t_1: t_2: t_3 & =1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2}) \end{aligned} $

As object return with smaller speed that means there is a loss of energy. This loss of energy occurs due to viscous drag of atmosphere. While object is moving up let viscous drag produces an deceleration ' $a$ '.

Then,

$ \begin{array}{ll} & v^2-u^2=2 a s \\ \Rightarrow \quad & 0-(10)^2=2(-(g+a)) h \\ \text { or } \quad & 100=2(g+a) h \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

While coming down,

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & 64-0=2(-(g-a))(-h) \\ \Rightarrow & 64=2(g-a) h \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array}\\ &\text { Dividing (i) and (ii) we get }\\ &\begin{aligned} & \frac{100}{64}=\frac{g+a}{g-a} \\ \Rightarrow \quad & 100 g-100 a=64 g+64 a \\ & 36 g=164 a \text { or } a=\frac{36}{164} g \end{aligned} \end{aligned} $

Substituting ' $a$ ' in either (i) or (ii) we get ' $h$.

From (ii)

$ \begin{aligned} & & 64 & =2\left(g-\frac{36}{164} g\right) h \\ \Rightarrow & & 32 & =\left(\frac{164-36}{164}\right) g h \Rightarrow \\ \Rightarrow & & h & =4.1 \mathrm{~m} \end{aligned} $

$u=72 \mathrm{~km} / \mathrm{h}=72 \times \frac{3}{18}=20 \mathrm{~m} / \mathrm{s}$

$ \begin{aligned} & a=-5 \mathrm{~m} / \mathrm{s}^2 \\ \therefore & \text { using }, v^2=u^2+2 a s \\ \Rightarrow & 0=20^2+2(-5) s \\ \Rightarrow & s=\frac{400}{10}=40 \mathrm{~m} \end{aligned} $

Distance travelled during reaction time

$ \begin{aligned} & =\text { Total distance }-s \\ & =50-40=10 \mathrm{~m} \end{aligned} $

If $t$ be the reaction time, then

$ \begin{aligned} & s=u t \\ & t=\frac{s}{u}=\frac{10}{20}=0.5 \mathrm{~s} \end{aligned} $

Using $v^2-u^2=2 g h$

We have for $h_1$ (see figure)

$ \begin{aligned} & 18^2-20^2=2(-10) \times h_1 \\ \Rightarrow \quad & h_1=3.8 \mathrm{~m} \end{aligned} $

Now for $h_2$,

$ \begin{aligned} & 0-18^2=2(-10) \times h_2 \\ \Rightarrow \quad & h_2=16.2 \mathrm{~m} \end{aligned} $

Maximum height attained

$ \begin{aligned} & =h_1+h_2=3.8+16.2 \\ & =20 \mathrm{~m} \end{aligned} $

The given situation is shown in figure.

Using $v=u-g t$, for motion point $A$ to $P$,

$ \begin{aligned} & v_1=v-g t_1 \\ & v_1=v-g x \end{aligned} $

At highest point $H, v_2=0$

$ \begin{aligned} \therefore & & v_2 & =v_1-g t_2 \\ & & 0 & =v_1-g t_2 \\ \Rightarrow & & t_2 & =\frac{v_1}{g}=\frac{v-g x}{g} \end{aligned} $

$\therefore$ Required time $=2 t_2$

$ =2\left(\frac{v-g x}{g}\right)=2\left(\frac{v}{g}-x\right) $

The height of both planes is $h = 20~\mathrm{m}$.

Finding Length of Each Inclined Plane:

For plane $A$, the length is $l_1 = \frac{h}{\sin 30^\circ}$.

Since $\sin 30^\circ = \frac{1}{2}$, we get $l_1 = \frac{20}{0.5} = 40~\mathrm{m}$.

For plane $B$, the length is $l_2 = \frac{h}{\sin 60^\circ}$.

Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we get $l_2 = \frac{20}{\frac{\sqrt{3}}{2}} = \frac{40}{\sqrt{3}}~\mathrm{m}$.

Finding Acceleration Along Each Plane:

Acceleration on an incline is $a = g \sin\theta$.

For plane $A$: $a_1 = 10 \times \sin 30^\circ = 10 \times 0.5 = 5~\mathrm{m/s}^2$.

For plane $B$: $a_2 = 10 \times \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}~\mathrm{m/s}^2$.

Finding the Time to Slide Down Each Plane:

The formula for distance with starting speed $0$ is $s = \frac{1}{2} a t^2$. Solving for $t$ gives $t = \sqrt{\frac{2s}{a}}$.

For plane $A$: $t_1 = \sqrt{\frac{2l_1}{a_1}} = \sqrt{\frac{2 \times 40}{5}} = \sqrt{16} = 4~\mathrm{s}$.

For plane $B$: $t_2 = \sqrt{\frac{2l_2}{a_2}} = \sqrt{\frac{2 \times \frac{40}{\sqrt{3}}}{5\sqrt{3}}}$.

Let's simplify $t_2$: $ t_2 = \sqrt{\frac{80/\sqrt{3}}{5\sqrt{3}}} = \sqrt{\frac{80}{5(\sqrt{3})^2}} = \sqrt{\frac{80}{5 \times 3}} = \sqrt{\frac{80}{15}} = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}}~\mathrm{s} $

Finding the Difference in Time:

$t_1 - t_2 = 4 - \frac{4}{\sqrt{3}}$

This can also be written as: $ 4 \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right)~\mathrm{s} $

$ \begin{aligned} \quad s & =t^3-6 t^2+18 t+9 \\ \therefore \quad v & =\frac{d s}{d t}=\frac{d}{d t}\left(t^3-6 t^2+18 t+9\right) \\ v & =3 t^2-12 t+18 \\ \frac{d v}{d t} & =6 t-12 \end{aligned} $

$v$ will be minimum if

$ \begin{aligned} \frac{d v}{d t} & =0 \\ 6 t-12 & =0 \\ t & =2 \mathrm{~s} \\ \therefore \quad v_{\min } & =3 \times 2^2-12 \times 2+18=6 \mathrm{~m} / \mathrm{s} \end{aligned} $

From the given graph,

At time $t=0 \mathrm{~s}$, initial displacement

$ x_i=80 \mathrm{~m} $

At time $t=10 \mathrm{~s}$, final displacement

$ x_f=60 \mathrm{~m} $

∴ Average velocity

$ \begin{aligned} & =\frac{x_f-x_i}{\Delta t}=\frac{60-80}{10} \\ & =\frac{-20}{-10}=-2 \mathrm{~m} / \mathrm{s} \end{aligned} $

∴ Magnitude of average velocity $=2 \mathrm{~m} / \mathrm{s}$

Step 1: Find acceleration $a$.

We use the formula for velocity with uniform acceleration: $v = u + a t$. Here, $u=0$ (because the body starts from rest), and after $n$ seconds, the velocity is $v$. So:

$ v = a n \implies a = \frac{v}{n} $

Step 2: Find displacement in the $n$th second ($S_n$).

The formula for how far an object moves during the $n$th second is:

$ S_n = u + \frac{a}{2}(2n-1) $

Since $u = 0$: $ S_n = \frac{a}{2}(2n-1) $

Step 3: Find displacement in the $(n-1)$th second ($S_{n-1}$).

Plug $(n-1)$ instead of $n$ into the formula:

$ \begin{aligned} S_{n-1} & = \frac{a}{2}(2(n-1)-1) \\ & = \frac{a}{2}(2n-3) \end{aligned} $

Step 4: Add the displacements for the $n$th and $(n-1)$th seconds.

$ S_n + S_{n-1} = \frac{a}{2}(2n-1) + \frac{a}{2}(2n-3) $

Combine the numbers in the brackets:

$ S_n + S_{n-1} = \frac{a}{2}[(2n-1) + (2n-3)] = \frac{a}{2}(4n-4) $

Step 5: Substitute $a = \frac{v}{n}$ from Step 1.

$ S_n + S_{n-1} = \frac{1}{2} \cdot \frac{v}{n} \cdot (4n-4) = \frac{v}{2n} (4n-4) $

Simplify:

$ S_n + S_{n-1} = \frac{v}{2n} \cdot 4(n-1) = \frac{2v}{n}(n-1) $

If $h$ be the height of the bridge above the water surface, then

$ \begin{aligned} h & =-u t+\frac{1}{2} g t^2 \\ & =-5 \times 3+\frac{1}{2} \times 10 \times 3^2 \\ & =-15+45=30 \mathrm{~m} \end{aligned} $

$ \frac{x}{2}=3 t_1 $

$ \begin{aligned} V_{\text {avg }} & =\frac{\text { total distance travelled }}{\text { total time taken }} \\ & =\frac{x}{t_1+t_2}=\frac{x}{\frac{x}{6}+\frac{x}{12}} \\ & =\frac{x}{\frac{3 x}{12}}=4 \mathrm{~m} / \mathrm{s} \end{aligned} $

Step 1: Write the formula for distance in the $n$-th second

The distance a freely falling object travels during the $n$-th second is given by: $ S_n = u + \frac{1}{2}g(2n - 1) $ Here, $u$ is the starting speed (which is $0$ if just falling), $g$ is gravity, and $n$ is the second we are looking at.

Step 2: Substitute the values

We know the speed at start $u=0$ and $g=10~\mathrm{ms}^{-2}$. The distance in the last but one second is $5$ m, so: $ 5 = 0 + \frac{1}{2}\times10(2n-1) $

Step 3: Solve for $n$

Simplify the equation: $ 5 = 5(2n-1) $ $ \Rightarrow \frac{5}{5} = 2n-1 $ $ \Rightarrow 1 = 2n-1 $ $ \Rightarrow 2n = 2 $ $ \Rightarrow n = 1 $

This means the last but one second is the 1st second of the fall.

$ \text { } \begin{aligned} \frac{h_2}{h_5} & =\frac{u+\frac{1}{2} g\left(2 n_1-1\right)}{u+\frac{1}{2} g\left(2 n_2-1\right)} \\ & =\frac{\frac{1}{2} g(2 \times 2-1)}{\frac{1}{2} g(2 \times 5-1)}=\frac{3}{9}=\frac{1}{3} \\ \therefore h_2: h_5 & =1: 3 \end{aligned} $

$ \begin{aligned} V_{\text {avg }} & =\frac{\text { total distance }}{\text { total time taken }} \\ & =\frac{0.4 \mathrm{~s}+0.6 \mathrm{~s}}{\frac{0.4 \mathrm{~s}}{v_1}+\frac{0.6 \mathrm{~s}}{v_2}} \\ & =\frac{\mathrm{s}}{\frac{2 \mathrm{~s}}{5} v_2+\frac{3 \mathrm{~s}}{5} v_1} \\ & =\frac{5 v_1 v_2}{3 v_1+2 v_2} \end{aligned} $

Let's analyze the given information before determining the distance between the two cars when they come to rest. Each car is traveling towards the other at a speed of $20 \, \text{m s}^{-1}$ and they both start braking when they are $300 \, \text{m}$ apart. The deceleration (negative acceleration) of each car is given as $2 \, \text{m s}^{-2}.$

To find the distance each car travels before coming to rest, we can use the kinematic equation that relates initial velocity ($v_i$), final velocity ($v_f$), acceleration ($a$), and distance ($d$), which is:

$v_f^2 = v_i^2 + 2ad$

Since the final velocity $v_f = 0$ (they come to rest), we can rearrange the equation to solve for $d$ (the distance each car travels before stopping):

$0 = v_i^2 + 2ad \Rightarrow d = -\frac{v_i^2}{2a}$

Plugging in the values for each car (noting that acceleration $a$ is negative because it is deceleration, so $a = -2 \, \text{m s}^{-2}$):

$d = -\frac{(20)^2}{2(-2)} = -\frac{400}{-4} = 100 \, \text{m}$

Each car travels $100 \, \text{m}$ before coming to rest. Since they both start $300 \, \text{m}$ apart and each travels $100 \, \text{m}$ towards the other, the total distance covered by both cars before stopping is $2 \times 100 \, \text{m} = 200 \, \text{m}$.

To find the distance between them when they come to rest, we subtract the total distance covered by both cars from the original distance between them:

$300 \, \text{m} - 200 \, \text{m} = 100 \, \text{m}$

So, the distance between the cars when they come to rest is $100 \, \text{m}$. Therefore, the correct option is:

Option B 100 m

Let's denote the total distance covered by the particle as $2d$, where $d$ is the distance for each half. To calculate the average speed, we need to find the total distance traveled and divide it by the total time taken.

For the first half of the journey, the particle covers the distance $d$ at a speed of $6 \, \text{m/s}$. The time taken for this part of the journey can be calculated using the formula $\text{time} = \frac{\text{distance}}{\text{speed}}$. So,

For the second half of the journey, the distance $d$ is further divided into two parts, each covered in equal time intervals. Given the speeds are $9 \, \text{m/s}$ and $15 \, \text{m/s}$ respectively, let's call the equal time intervals $t$. The distances covered in these intervals can be found by $\text{distance} = \text{speed} \times \text{time}$.

For the part covered at $9 \, \text{m/s}$:

For the part covered at $15 \, \text{m/s}$:

Since these two parts together make up the second half of the journey,

This gives us $24t = d$, and from this, we can find $t = \frac{d}{24}$.

The total time for the second half of the journey is the sum of the times for the two parts, which are equal ($t$ each), so the total time for the second half is $2t$. Since $t = \frac{d}{24}$,

The total time taken for the entire journey is the sum of the times for the first and second halves:

The total distance is $2d$, and the total time is $\frac{d}{4}$. Therefore, the average speed is calculated as:

Thus, the correct answer is Option D: 8 m/s.

To solve this problem, we'll use the equations of motion under constant acceleration due to gravity. Let's define:

$ h $: Height of the tower

$ u $: Initial speed of projection (same magnitude in both cases)

$ g $: Acceleration due to gravity (positive downward)

$ t_1 $: Time taken to reach the ground when projected upwards

$ t_2 $: Time taken to reach the ground when projected downwards

$ t $: Time taken to reach the ground when simply dropped

Case 1: Projectile Thrown Upwards

When the body is projected upwards from the top of the tower, its initial velocity is $ -u $ (since upward direction is negative), and it reaches the ground in time $ t_1 $. The equation of motion is:

$ h = -u t_1 + \frac{1}{2} g t_1^2 $

Case 2: Projectile Thrown Downwards

When the body is projected downwards from the top of the tower, its initial velocity is $ u $, and it reaches the ground in time $ t_2 $. The equation is:

$ h = u t_2 + \frac{1}{2} g t_2^2 $

Case 3: Body Dropped

When the body is simply dropped, its initial velocity is $ 0 $, and it reaches the ground in time $ t $:

$ h = \frac{1}{2} g t^2 $

Step 1: Equate the Heights

From cases 1 and 2, equate the expressions for $ h $:

$ -u t_1 + \frac{1}{2} g t_1^2 = u t_2 + \frac{1}{2} g t_2^2 $

Step 2: Solve for $ u $

Simplify the equation:

$ -u t_1 - u t_2 = \frac{1}{2} g t_2^2 - \frac{1}{2} g t_1^2 $

$ -u (t_1 + t_2) = \frac{1}{2} g (t_2^2 - t_1^2) $

$ -u (t_1 + t_2) = \frac{1}{2} g (t_2 - t_1)(t_2 + t_1) $

Divide both sides by $ t_1 + t_2 $:

$ -u = \frac{1}{2} g (t_2 - t_1) $

$ u = \frac{1}{2} g (t_1 - t_2) $

Step 3: Express $ h $ in Terms of $ t_1 $ and $ t_2 $

Substitute $ u $ back into one of the equations for $ h $:

$ h = -\left( \frac{1}{2} g (t_1 - t_2) \right) t_1 + \frac{1}{2} g t_1^2 $

Simplify:

$ h = \frac{1}{2} g t_1 t_2 $

Step 4: Equate the Height for the Dropped Case

From the dropped case:

$ h = \frac{1}{2} g t^2 $

Set the two expressions for $ h $ equal to each other:

$ \frac{1}{2} g t^2 = \frac{1}{2} g t_1 t_2 $

$ t^2 = t_1 t_2 $

Step 5: Solve for $ t $

$ t = \sqrt{t_1 t_2} $

Conclusion:

The time required for the body to reach the ground when dropped is the geometric mean of $ t_1 $ and $ t_2 $:

Answer: Option B

$ t = \sqrt{t_1 \, t_2} $

To find the average speed of the train for the duration of the journey, we need to know the total distance covered by the train and the total time taken.

The train accelerates uniformly to a speed of $80 \, \mathrm{km/h}$ over time $t$, and then moves at this constant speed for $3t$. The average speed can be calculated using the formula:

$\text{Average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}}$

Step 1: Calculate the distance covered during acceleration

The distance covered while the train is accelerating can be found using the formula for the distance travelled under uniform acceleration:

$d_1 = \frac{1}{2} at^2$

Where:

• $d_1$ is the distance covered during acceleration.
• $a$ is the acceleration (we don't have a direct value for this, but we can work with the given information).
• $t$ is the time.

However, to proceed with the calculation without the acceleration ($a$), we recognize that the formula directly correlates to distance but requires knowledge of acceleration. Instead, let's think in terms of the final speed and time, given that the train reaches $80 \, \mathrm{km/h}$ (or $\frac{80}{3.6} = 22.22 \, \mathrm{m/s}$) in time $t$.

Using the relationship between velocity, time, and distance, since the acceleration is uniform, we can use:

$d_1 = v \times t_1 - \frac{1}{2} a t^2$

Given that the initial speed $u = 0$ and final speed $v = 80 \, \mathrm{km/h}$, converting the speed to meters per second (since our time is likely in seconds) gives us $22.22 \, \mathrm{m/s}$. But without directly calculating acceleration, we simplify using average speed for the acceleration phase because it starts from rest and reaches $v$.

The average speed during acceleration, \(v_{avg} = \frac{u + v}{2} = \frac{0 + 80}{2} = 40 \, \mathrm{km/h}$.

Thus, the distance $d_1 = v_{avg} \times t = 40 \, \mathrm{km/h} \times t$.

Step 2: Calculate the distance covered at constant speed

The distance covered at a constant speed is easier to calculate:

$d_2 = v \times t_2 = 80 \, \mathrm{km/h} \times 3t$

Step 3: Calculate the total distance and the total time

The total distance ($D$) covered is the sum of $d_1$ and $d_2$:

$D = d_1 + d_2 = 40t + 240t = 280t \, \mathrm{km}$

The total time ($T$) taken is $t + 3t = 4t$.

Step 4: Calculate the average speed

Substitute the values of $D$ and $T$ in the formula of average speed:

$\text{Average speed} = \frac{280t}{4t}$

This simplifies to $70 \, \mathrm{km/h}$.

So, the average speed of the train for this duration of the journey is $70 \, \mathrm{km/h}$, which matches with Option A.

The distance covered by a body in the $n^{\text{th}}$ second can be found using the equation:

$S_{n} = u + \dfrac{1}{2}a(2n-1)$

where,

• $S_{n}$ is the distance covered in the $n^{\text{th}}$ second,
• $u$ is the initial velocity,
• $a$ is the acceleration, and
• $n$ is the nth second.

The distance covered in the $n^{\text{th}}$ second is given as $102.5 \, \text{m}$, so we have:

$102.5 = u + \dfrac{1}{2}a(2n-1)$ ---- (1)

For the $(n + 2)^{\text{th}}$ second, the distance covered is:

$S_{n+2} = u + \dfrac{1}{2}a(2(n+2)-1)$

Substituting $n + 2$ in place of $n$, we get:

$115.0 = u + \dfrac{1}{2}a(2n+3)$ ---- (2)

Subtracting equation (1) from equation (2), we get:

$115.0 - 102.5 = \dfrac{1}{2}a(2n + 3 - 2n + 1)$

$12.5 = \dfrac{1}{2}a(4)$

So, solving for $a$ gives:

$a = \dfrac{12.5 \times 2}{4} = \dfrac{25}{4} = 6.25 \, \text{m/s}^2$

Therefore, the acceleration of the body is:

$6.25 \, \text{m/s}^2$

Which corresponds to Option A: $6.25 \, \text{m/s}^2$.

To find the velocity of Train B with respect to Train A, we have to subtract the velocity of Train A from the velocity of Train B, keeping in mind that they are moving in opposite directions. Since they are moving in opposite directions, the relative velocity is calculated by adding their magnitudes when converting into the same unit, which in this case is meters per second.

First, let's convert the speeds from km/h to m/s by multiplying by the conversion factor $ \frac{1000 \text{ m/km}}{3600 \text{ s/h}} = \frac{5}{18} \text{ m/s} $.

For Train A:

For Train B:

To find the velocity of B relative to A ($ v_{B/A} $), we consider the direction: Train B is moving towards the south and Train A is moving towards the north. Therefore, relative to Train A, Train B is moving even faster towards the south, we calculate:

Since Train B is moving towards the south and Train A towards the north, we take the southward direction as negative in our coordinate system for this calculation. That means the velocity of B with respect to A is:

Next, we calculate the velocity of the ground with respect to Train B ($ v_{ground/B} $). The ground is stationary, thus it has a velocity of 0 m/s in any direction. The velocity of an object with respect to another object moving is just the opposite of the second object's velocity. Thus:

However, because we are considering the southward direction as negative, the negative of a southward velocity is a northward velocity. Hence we get:

So the velocity of Train B with respect to Train A is -50 m/s, and the velocity of the ground with respect to Train B is 30 m/s. Therefore, the correct answer is:

Option A: -50 m/s and -30 m/s. (Incorrect, because the velocity of ground with respect to B is positive in our chosen coordinate system)

Option B: -50 m/s and 30 m/s. (Correct)

Option C: -30 m/s and 50 m/s. (Incorrect)

Option D: 50 m/s and -30 m/s. (Incorrect)

Thus, the correct answer is Option B: -50 m/s and 30 m/s.

The relationship between time ($t$) and distance ($x$) is given by $t = \alpha x^2 + \beta x$, where $\alpha$ and $\beta$ are constants. To find the relation between acceleration ($a$) and velocity ($v$), we can follow these steps:

First, we differentiate the given equation with respect to time:

$\begin{aligned} & t = \alpha x^2 + \beta x \quad \text{(differentiating with respect to time)} \\ & \frac{dt}{dx} = 2\alpha x + \beta \\ & \frac{1}{v} = 2\alpha x + \beta \\ \end{aligned}$

Next, we differentiate again with respect to time to find the acceleration:

$\begin{aligned} & -\frac{1}{v^2} \frac{dv}{dt} = 2\alpha \frac{dx}{dt} \\ & \text{Since} \quad \frac{dx}{dt} = v, \quad \text{we have:} \\ & -\frac{1}{v^2} \frac{dv}{dt} = 2\alpha v \\ & \frac{dv}{dt} = -2\alpha v^3 \\ & \text{Therefore,} \quad a = -2\alpha v^3 \end{aligned}$

The position equation is given by:

$x = t^3 - 6t^2 + 20t + 15$

First, compute the velocity $v$ by differentiating the position function with respect to time:

$ \frac{d x}{d t} = v = 3t^2 - 12t + 20 $

Next, compute the acceleration $a$ by differentiating the velocity function with respect to time:

$ \frac{d v}{d t} = a = 6t - 12 $

We need to find the time $t$ when the acceleration is zero:

$ 6t - 12 = 0 $

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

Now, find the velocity at $t = 2 \, \mathrm{sec}$:

$ v = 3(2)^2 - 12(2) + 20 $

$ v = 8 \, \mathrm{m/s} $

Let's denote the constant acceleration with which the body moves as $a$. We know that the displacement $S$ covered by a body starting from rest under a constant acceleration $a$ in time $t$ is given by the equation of motion: $S = \frac{1}{2} a t^2$.

Considering the displacement $\mathrm{S}_1$ in first $(p-1)$ seconds, we apply the equation of motion:

Similarly, for the displacement $\mathrm{S}_2$ in first $p$ seconds:

To find out the total time it will take to cover the displacement $\mathrm{S}_1+\mathrm{S}_2$, we first find the sum of these two displacements:

Let's simplify this:

If we consider the total displacement $\mathrm{S}_1+\mathrm{S}_2$ is to be covered in a time $t$ seconds from rest, we should set this equal to the equation of motion:

Equating the two equations:

Since $a \neq 0$, we can simplify by dividing both sides by $ \frac{1}{2} a$:

To find $t$, we take the square root of both sides:

Therefore, the time taken to cover the displacement $\mathrm{S}_1+\mathrm{S}_2$ will be:

Hence, the correct option would be: