\[ S= \frac{1}{2}at^{2}\]

As, S is proportional to t graph is a parabola

Area bounded by v-t graph represents displacement. Here.

Area = (1/2)×4×8= 16

So, Distance travelled is 16 m.

Area of acceleration time graph represents change in velocity. As acceleration is always positive speed is always increasing.

Area = (1/2)× 10× 11 = 55

So, max. speed of 55 m/s

Using relative speed, let’s re-calculate with the correct setup:

1. Acceleration of the bus: \( a = 1 \, \text{m/s}^2 \)

2. Initial distance behind the bus: \( d = 48 \, \text{m} \)

3. Speed of the man: \( v_m = 10 \, \text{m/s} \)

4. The equation of motion for the bus after \( t \) seconds:
\[
d_b = \frac{1}{2} a t^2 = \frac{1}{2} \cdot 1 \cdot t^2 = \frac{1}{2} t^2
\]

5. Distance traveled by the man:
\[
d_m = v_m \cdot t = 10t
\]

6. Setting the distances equal considering the man starts 48 m behind:
\[
10t = \frac{1}{2} t^2 + 48
\]
\[
\frac{1}{2} t^2 - 10t + 48 = 0
\]

7. Multiplying by 2:
\[
t^2 - 20t + 96 = 0
\]

8. Using the quadratic formula:
\[
t = \frac{20 \pm \sqrt{20^2 - 4 \cdot 96}}{2}
\]
\[
t = \frac{20 \pm \sqrt{400 - 384}}{2}
\]
\[
t = \frac{20 \pm 4}{2}
\]
\[
t = 12 \, \text{s} \quad \text{or} \quad t = 8 \, \text{s}
\]

 the man will catch the bus is 8 seconds. 

Speed at mid point is given by

\[ V_{mid}=\sqrt{\frac{V_{1}^{2}+V_{2}^{2}}{2}} \]

1. Time Ratios: The time ratios are given as
   $1:8:1$ 

   $1:8:1$. So, if the total time is $t$ 

   $t$, the train spends $\frac{t}{10}$ 

   10t​ accelerating, $\frac{8t}{10}$ 

   108t​ at constant velocity, and $\frac{t}{10}$ 

   10t​ decelerating.

2. Velocity-Time Graph:
   • The graph forms a trapezium.
   • The train accelerates linearly from 0 to 60 km/h, holds constant at 60 km/h, and then decelerates back to 0.

   Key points:

   • The area under this graph represents the total distance traveled.
   • The height (maximum velocity) = 60 km/h.
   • The time intervals are in the ratio
     $1:8:1$ 

     $1:8:1$.

3. Average Speed:
   The area of the trapezium is given by: 

   $$\text{Area} = \frac{1}{2} \times (t_{\text{total}}) \times (\text{initial velocity} + \text{final velocity})$$Area=21​×(ttotal​)×(initial velocity+final velocity)For the constant velocity portion:

    

   $$\text{Average speed} = \frac{60 \times (1 + 8 + 1)}{10} = 54 \, \text{km/h}$$Average speed=1060×(1+8+1)​=54km/h

Thus, the average speed is 54 km/h.

To find the resultant velocity using unit vectors:

- Initial velocity: \( \vec{u} = 0.4 \, \hat{i} \, \text{m/s} \)
- Acceleration: \( \vec{a} = 0.15 \, \hat{j} \, \text{m/s}^2 \)
- Time: \( t = 2 \, \text{seconds} \)

Using the equation of motion \( \vec{v} = \vec{u} + \vec{a}t \):

- In the \( x \)-direction: \( v_x = 0.4 \, \text{m/s} \)
- In the \( y \)-direction: \( v_y = 0 + 0.15 \times 2 = 0.3 \, \text{m/s} \)

The resultant velocity is:

\[
|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{(0.4)^2 + (0.3)^2} = \sqrt{0.25} = 0.5 \, \text{m/s}
\]

Thus, the resultant velocity is 0.5 m/s

The time taken by an object to fall from a height is given by the equation:

\[
t = \sqrt{\frac{2h}{g}}
\]

For mass \( m_a \) dropped from height \( a \), the time taken is:

\[
t_a = \sqrt{\frac{2a}{g}}
\]

For mass \( m_b \) dropped from height \( b \), the time taken is:

\[
t_b = \sqrt{\frac{2b}{g}}
\]

The ratio of time taken by the two bodies is:

\[
\frac{t_a}{t_b} = \frac{\sqrt{\frac{2a}{g}}}{\sqrt{\frac{2b}{g}}} = \sqrt{\frac{a}{b}}
\]

So, the ratio of the time taken is:

\[
\frac{t_a}{t_b} = \sqrt{\frac{a}{b}}
\]

Let the initial velocity of the ball be \( u \), and let \( v \) be the velocity of the ball at a distance \( h \) above the tower. Using the equation of motion:

\[
v^2 = u^2 - 2gh
\]

At a distance \( h \) below the tower, the velocity is doubled, so:

\[
(2v)^2 = u^2 + 2gh
\]

Simplifying these:

\[
4v^2 = u^2 + 2gh
\]

Substitute \( v^2 = u^2 - 2gh \) from the first equation:

\[
4(u^2 - 2gh) = u^2 + 2gh
\]

Expanding and solving:

\[
4u^2 - 8gh = u^2 + 2gh
\]

\[
3u^2 = 10gh
\]

\[
u^2 = \frac{10gh}{3}
\]

The maximum height \( H \) from the top of the tower is given by:

\[
H = \frac{u^2}{2g} = \frac{10gh}{6g} = \frac{5h}{3}
\]

Thus, the maximum height attained by the ball from the top of the tower is \( \frac{5h}{3} \).

From Galileo's ratio of odd number distances travelled in consecutive seconds are in the order of 1:3:5:7... Here distance travelled in 3rd second is 5/9th of total distance total time of flight is 3 second.

\[ s= \frac{1}{2}gt^{2}= \frac{1}{2}\times 10\times 3^{2}= 45 m \]