Given acceleration:

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

The velocity after time \(t\) is:

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

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

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

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