Given,

$x(t) = x_{0}(1 - e^{-γt})$

$v(t) = \dfrac{dx(t)}{dt} = x_{0}γe^{-γt}$

$a(t) = \dfrac{dv(t)}{dt} = - x_{0}γ^{2}e^{-γt}$

(a) When $t = 0$; $x(t) = x_{0}(1 - e^{0}) = x_{0}(1 - 1) = 0$

$x(t = 0) = x_{0}γe^{0} = x_{0}γ(1) = γx_{0}$

(b) $x(t)$ is maximum when $t = ∞$      $[x(t)]_{max} = x_{0}$

$x(t)$ is minimum when $t = 0$     $[x(t)]_{min} = 0$

$v(t)$ is maximum when $t = 0$; $v(0) = x_{0}γ$

$v(t)$ is minimum when $t = ∞$; $v(∞) = 0$

$a(t)$ is maximum when $t = ∞$; $a(∞) = 0$

$a(t)$ is minimum when $t = 0$; $a(0) = - x_{0}γ^{2}$

Note We should be careful about the nature of variation of the curve and maximum and minimum value will be decided accordingly.

To determine the total distance covered by the bird, let's proceed step-by-step.

First, we need to calculate the time it takes for the two cars to meet. The two cars are moving towards each other, so their relative speed is the sum of their individual speeds.

The speed of the first car is 18 km/h and the speed of the second car is 27 km/h. Therefore, their relative speed is:

$$18 \, \text{km/h} + 27 \, \text{km/h} = 45 \, \text{km/h}$$

The initial distance between the two cars is 36 km. To find the time it takes for the cars to meet, we use the formula:

$$\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}}$$

Substituting the provided values, we get:

$$\text{Time} = \frac{36 \, \text{km}}{45 \, \text{km/h}} = \frac{4}{5} \, \text{hours} = 0.8 \, \text{hours}$$

Now, we know that the bird is flying at a speed of 36 km/h. To find the total distance covered by the bird while the two cars meet, we use the formula:

$$\text{Distance} = \text{Speed} \times \text{Time}$$

Substituting the speed of the bird and the time calculated, we get:

$$\text{Distance} = 36 \, \text{km/h} \times 0.8 \, \text{hours} = 28.8 \, \text{km}$$

Thus, the total distance covered by the bird is 28.8 kilometers.

Given, horizontal speed of the man ($u_x$) = 9 m/s

Horizontal distance between the two buildings = 10 m

Height difference between the two buildings = 9 m

and $g = 10 \text{ m/s}^2$

Let the man jumps from point $A$ and land on the roof of the next building at point $B$.

Taking motion in vertical direction,

$y = ut + \frac{1}{2} at^2$

$9 = 0 \times t + \frac{1}{2} \times 10 \times t^2$

$9 = 5t^2$

or $t = \sqrt{ \frac{9}{5}} = \frac{3}{\sqrt{5}}$

Therefore, horizontal distance travelled = $u_x \times t = 9 \times \frac{3}{\sqrt{5}} = \frac{27}{\sqrt{5}} \approx 12$ m

Horizontal distance travelled by the man is greater than 10 m, therefore, he will land on the next building.

For the ball dropped from the building, $u_1 = 0, u_2 = 40 \text{ m/s}$
Velocity of the dropped ball after time $t$,

$v_1 = u_1 + gt$

$v_1 = gt$ (downward)

For the ball thrown up, $u_2 = 40 \text{ m/s}$

Velocity of the ball after time $t$:

$v_2 = u_2 - gt$

$v_2 = (40 - gt) $ (upward)

∴ Relative velocity of one ball with respect to another ball:

$v_1 - v_2$

$= gt - [40 - gt]$

$= 40 \text{ m/s}$

Note: When we are applying equations for rectilinear motion we should carefully put up the signs for the physical quantities.

Given, initial velocity $= v_0$

Let the distance traveled in time $t = x_0$.

For the graph:

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

where, $v$ is velocity, and $x$ is the displacement at any instant of time $t$.

From Eq. ($i$):

$v_0 - v = \frac{v_0}{x_0} x$

⇒ $v = -\frac{v_0}{x_0} x + v_0$

We know that

$$ \begin{aligned} \text { Acceleration } a & =\frac{d v}{d t}=\frac{-v_0}{x_0} \frac{d x}{d t}+0 \\\\ \Rightarrow a & =\frac{-v_0}{x_0}(v) \\\\ & =\frac{-v_0}{x_0}\left(\frac{-v_0}{x_0} x+v_0\right) \\\\ & =\frac{v_0{ }^2}{x_0{ }^2} x-\frac{v_0{ }^2}{x_0} \end{aligned} $$

Graph of a versus $x$ is given above.