Given:

\[
x = at + bt^2 - ct^3
\]

1. Velocity:

\[
v = \frac{dx}{dt} = a + 2bt - 3ct^2
\]

2.Acceleration:

\[
a = \frac{dv}{dt} = 2b - 6ct
\]

3. Set acceleration to zero:

\[
2b - 6ct = 0 ; t = \frac{b}{3c}
\]

4. Velocity at \(t = \frac{b}{3c}\)

\[
v = a + 2b\left(\frac{b}{3c}\right) - 3c\left(\frac{b}{3c}\right)^2
\]

\[
= a + \frac{2b^2}{3c} - \frac{b^2}{3c} = a + \frac{b^2}{3c}
\]

Given:

- Initial velocity: \(\mathbf{u} = 3\hat{i} + 4\hat{j}\)
- Acceleration: \(\mathbf{a} = 0.4\hat{i} + 0.3\hat{j}\)

Calculate final velocity after 10 s:

\[
\mathbf{v} = \mathbf{u} + \mathbf{a}t = (3\hat{i} + 4\hat{j}) + (0.4\hat{i} + 0.3\hat{j}) \cdot 10
\]

\[
\mathbf{v} = 3\hat{i} + 4\hat{j} + (4\hat{i} + 3\hat{j}) = (3 + 4)\hat{i} + (4 + 3)\hat{j} = 7\hat{i} + 7\hat{j}
\]

Calculate speed:

\[
\text{Speed} = |\mathbf{v}| = \sqrt{(7)^2 + (7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2} \, \text{m/s}
\]

Given the acceleration:

\[
a = 6t + 5 \, \text{m/s}^2
\]

1. Integrate acceleration to find velocity \(v\):

\[
v = \int a \, dt = \int (6t + 5) \, dt = 3t^2 + 5t + C
\]

Since it starts from rest, \(C = 0\):

\[
v = 3t^2 + 5t
\]

2. Integrate velocity to find displacement \(x\):

\[
x = \int v \, dt = \int (3t^2 + 5t) \, dt = t^3 + \frac{5}{2}t^2 + D
\]

Starting from the origin gives \(D = 0\):

\[
x = t^3 + \frac{5}{2}t^2
\]

3. Calculate displacement at \(t = 2\) s:

\[
x(2) = (2)^3 + \frac{5}{2}(2)^2 = 8 + \frac{5}{2} \cdot 4 = 8 + 10 = 18 \, \text{m}
\]

Thus, the distance covered in 2 seconds is: 18 m

Given the velocity:

$$v=\sqrt{180-16x}\text{m/s}$$

To find the acceleration, use the chain rule:

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

Differentiate

$v$

v with respect to

$x$

x:

$$\frac{dv}{dx} = \frac{-8}{\sqrt{180 - 16x}}$$

​

Now, substitute

$v = \sqrt{180 - 16x}$

$$a=-8{\text{m/s}}^2$$

Thus, the acceleration is

$a = -8 \, \text{m/s}^2$