Solution:

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 \(t = \frac{b}{3c}\)

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

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

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Solution:

Given acceleration:

\[
a = \alpha t + \beta
\]

The velocity after time \(t\) is:

\[
v(t) = \frac{\alpha t^2}{2} + \beta t
\]

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Solution:

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:

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

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Solution:

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

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Solution:

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

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Solution:

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:

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

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Solution:

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 \(x = \frac{\alpha^2 t^2}{4}\).

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Solution:

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 $v$ with respect to $x$:

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

Now, substitute $v = \sqrt{180 - 16x}$

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

Thus, the acceleration is $a = -8 \, \text{m/s}^2$

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Solution:

To find the acceleration, we need to compute the second derivatives of $x$ and $y$ with respect to $t$.

1. $x = 7t + 4t^2$
   
   • First derivative (velocity in x-direction): $$\frac{dx}{dt}=7+8t$$
   • Second derivative (acceleration in x-direction): $$\frac{d^2x}{dt^2} = 8 \, \text{m/s}^2$$
     
2. $y = 5t$
   • First derivative (velocity in y-direction): $$\frac{dy}{dt}=5$$
   • Second derivative (acceleration in y-direction): $$\frac{d^{2}y}{d{t}^2}=0{\text{m/s}}^2$$

Now, the total acceleration is given by:

$$a=\sqrt{(a_x{)}^2+(a_y{)}^2}=\sqrt{(8{)}^2+(0{)}^2}=8{\text{m/s}}^2$$

Thus, the acceleration of the particle at $t = 5 \, \text{s}$ is $8 \, \text{m/s}^2$

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