Motion in a Straight Line

As the meeting point lies $100 \mathrm{~m}$ above ground, displacement of ball will be $80 \mathrm{~m}$. For ball $A$

As ball $B$ is thrown after 2 seconds after release of $A$.

Thus, time available for ball $B$ is 2 seconds to cover a distance of $80 \mathrm{~m}$.

Let speed be ' $u$ ' $\mathrm{m} / \mathrm{s}, t_2=4-2=2 \mathrm{~s}, \mathrm{~S}=80 \mathrm{~m}$, $a=+g=+10 \mathrm{~m} / \mathrm{s}^2$

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: $h = -u t_1 + \dfrac{1}{2} g t_1^2$ 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: $h = u t_2 + \dfrac{1}{2} g t_2^2$ Case 3: Body Dropped When the body is simply dropped, its initial velocity is $0$, and it reaches the ground in time $t$: $h = \dfrac{1}{2} g t^2$ Step 1: Equate the Heights From cases 1 and 2, equate the expressions for $h$: $-u t_1 + \dfrac{1}{2} g t_1^2 = u t_2 + \dfrac{1}{2} g t_2^2$ Step 2: Solve for $u$ Simplify the equation: $-u t_1 - u t_2 = \dfrac{1}{2} g t_2^2 - \dfrac{1}{2} g t_1^2$ $-u (t_1 + t_2) = \dfrac{1}{2} g (t_2^2 - t_1^2)$ $-u (t_1 + t_2) = \dfrac{1}{2} g (t_2 - t_1)(t_2 + t_1)$ Divide both sides by $t_1 + t_2$: $-u = \dfrac{1}{2} g (t_2 - t_1)$ $u = \dfrac{1}{2} g (t_1 - t_2)$ Step 3: Express $h$ in Terms of $t_1$ and $t_2$ Substitute $u$ back into one of the equations for $h$: $h = -\left( \dfrac{1}{2} g (t_1 - t_2) \right) t_1 + \dfrac{1}{2} g t_1^2$ Simplify: $h = \dfrac{1}{2} g t_1 t_2$ Step 4: Equate the Height for the Dropped Case From the dropped case: $h = \dfrac{1}{2} g t^2$ Set the two expressions for $h$ equal to each other: $\dfrac{1}{2} g t^2 = \dfrac{1}{2} g t_1 t_2$ $t^2 = t_1 t_2$ 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 $t = \sqrt{t_1 \, t_2}$

Given

On integrating,

Let's denote the total distance covered by the particle as

, where

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

at a speed of

. The time taken for this part of the journey can be calculated using the formula

. So,

For the second half of the journey, the distance

is further divided into two parts, each covered in equal time intervals. Given the speeds are

and

respectively, let's call the equal time intervals

. The distances covered in these intervals can be found by

. For the part covered at

:

For the part covered at

:

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

This gives us

, and from this, we can find

. The total time for the second half of the journey is the sum of the times for the two parts, which are equal (

each), so the total time for the second half is

. Since

,

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

The total distance is

, and the total time is

. Therefore, the average speed is calculated as:

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

Given, $t=\sqrt{x}+4$, Squaring on both

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

. 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.

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: $H = \dfrac{1}{2}\,g\,t^2$ Substituting the given values: $19.6 = 4.9\,t^2$ Solving for $t$: $t^2 = \dfrac{19.6}{4.9}$ $t^2 = 4$ $t = 2 \, \text{sec}$ 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: $9 \, \text{km/h} = 9 \times \dfrac{1000}{3600} \, \text{m/s} = 2.5 \, \text{m/s}$ 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.

To find the length of train B, we first need to determine the relative velocity between train A and train B.

Since they are moving in opposite directions, their velocities add up:

Now, we need to convert this relative velocity to meters per second:

The passenger in train A observes train B for 8 seconds. To find the length of train B, we can use the formula:

So, the length of train B is 320 meters.

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 $x$: $\dfrac{dt}{dx} = 2mx + n$ Since velocity $v$ is defined as $\dfrac{dx}{dt}$, we can write: $\dfrac{1}{v} = \dfrac{dt}{dx} = 2mx + n$ Thus, $v = \dfrac{1}{2mx + n}$ Next, to find the acceleration $a$ (which is the derivative of velocity with respect to time), we start with the chain rule: $\dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt}$ Since $v = \dfrac{dx}{dt}$, substituting $x$ gives: $\dfrac{dx}{dt} = v$ Rewrite it: $\dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot v$ Differentiate $v$ with respect to $x$: $v = (2mx + n)^{-1}$ $\dfrac{dv}{dx} = -\dfrac{2m}{(2mx + n)^2}$ Then, $\dfrac{dv}{dt} = -\dfrac{2m}{(2mx + n)^2} \cdot v$ Substitute $v = \dfrac{1}{2mx + n}$ into the expression: $\dfrac{dv}{dt} = -\dfrac{2m}{(2mx + n)^2} \cdot \dfrac{1}{2mx + n}$ Simplify the expression: $\dfrac{dv}{dt} = -2m \left( \dfrac{1}{2mx + n} \right)^3$ $a = -2m v^3$ So, the retardation (negative acceleration) is: $a = -2m v^3$

Velocity of man with respect to ground

=

+

= -1.8 + 36 Velocity of man w.r.t. B

=

-

= –1.8 + 36 – (–72) = 106.2 km/hr = 29.5 m/s