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 position vector:

\[
\mathbf{r} = (3t)\hat{i} - (t^2)\hat{j} + 4\hat{k}
\]

To find the velocity, differentiate the position vector with respect to time \(t\):

\[
\mathbf{v} = \frac{d\mathbf{r}}{dt} = \left( \frac{d}{dt}(3t) \hat{i} + \frac{d}{dt}(-t^2) \hat{j} + \frac{d}{dt}(4) \hat{k} \right)
\]

Calculating the derivatives:

\[
\mathbf{v} = (3)\hat{i} - (2t)\hat{j} + (0)\hat{k} = 3\hat{i} - 2t\hat{j}
\]

Now, substitute \(t = 5\) s:

\[
\mathbf{v}(5) = 3\hat{i} - 2(5)\hat{j} = 3\hat{i} - 10\hat{j}
\]

Calculate the magnitude of the velocity:

\[
|\mathbf{v}| = \sqrt{(3)^2 + (-10)^2} = \sqrt{9 + 100} = \sqrt{109}
\]

Thus, the magnitude of velocity after 5 seconds is: 10.44 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 function:

\[
v = v_0 + gt + ft^2
\]

We can find displacement by integrating the velocity function with respect to time. The displacement \(x(t)\) is the integral of velocity:

\[
x(t) = \int (v_0 + gt + ft^2) \, dt
\]

Integrating:

\[
x(t) = v_0 t + \frac{gt^2}{2} + \frac{ft^3}{3} + C
\]

Since \(x = 0\) at \(t = 0\), we know \(C = 0\). Therefore, the position function is:

\[
x(t) = v_0 t + \frac{gt^2}{2} + \frac{ft^3}{3}
\]

Now, for \(t = 1\):

\[
x(1) = v_0(1) + \frac{g(1)^2}{2} + \frac{f(1)^3}{3}
\]

Simplifying:

\[
x(1) = v_0 + \frac{g}{2} + \frac{f}{3}
\]

Thus, the displacement after unit time \(t = 1\) is:

\[
x(1) = v_0 + \frac{g}{2} + \frac{f}{3}
\]

Given that the velocity \(v = \alpha \sqrt{x}\), we need to find the displacement \(x\) as a function of time \(t\).

1. \( v = \frac{dx}{dt} = \alpha \sqrt{x} \)

2. Rearranging and separating variables:
\[
\frac{dx}{\sqrt{x}} = \alpha \, dt
\]

3. Integrate both sides:
\[
2\sqrt{x} = \alpha t
\]

4. Solving for (x):
\[ x = \frac{\alpha^2 t^2}{4} \]

Thus, the displacement of the particle varies with time as \(x = \frac{\alpha^2 t^2}{4}\).

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

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

Slope of displacement time graph represents velocity.

V_A= tan 30 ⁰
V_B= tan 45 ⁰

So, V_A/V_B=tan 30 ⁰/tan 45 ⁰= 1/√3

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

As s is proportional to square of t. s-t graph is a parabola.

Slope of v-t graph represent acceleration. when object is thrown upward its acceleration is -g for the entire journey. So, slope is -g.

\[ v = u + at ; as, u = 0 ; v = at\]

As, v is proportional to t ; graph of v-t is a straight line.