A particle moves in such a manner that `x=At, y=Bt^(3)-2t, z=ct^(2)-4t`, where x, y and z are measured in meters and t is measured in seconds, and A, B and C are unknown constants. Give that the velocity of the particle at `t=2s` is `vec(v)=((dvec(r))/(dt))=3hat(i)+22hat(j) m//s`, determine the velocity of the particle at `t=4s`.

Explanation

To solve the problem, we need to determine the velocity of a particle at \( t = 4 \) seconds given its position equations and the velocity at \( t = 2 \) seconds.

### Step-by-Step Solution:

1. **Write the position vector**: The position vector \( \vec{r} \) of the particle is given by:
\[ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \]
where
\[ x = At, \quad y = Bt^3 - 2t, \quad z = Ct^2 - 4t \]
Thus, we can write:
\[ \vec{r} = At \hat{i} + (Bt^3 - 2t) \hat{j} + (Ct^2 - 4t) \hat{k} \]

2. **Differentiate the position vector to find the velocity vector**: The velocity vector \( \vec{v} \) is given by the derivative of the position vector with respect to time \( t \):
\[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(At) \hat{i} + \frac{d}{dt}(Bt^3 - 2t) \hat{j} + \frac{d}{dt}(Ct^2 - 4t) \hat{k} \]
Calculating the derivatives:
\[ \vec{v} = A \hat{i} + (3Bt^2 - 2) \hat{j} + (2Ct - 4) \hat{k} \]

3. **Substitute \( t = 2 \) seconds to find constants \( A, B, C \)**: We know that at \( t = 2 \) seconds, \( \vec{v} = 3 \hat{i} + 22 \hat{j} \).
\[ \vec{v} = A \hat{i} + (3B(2^2) - 2) \hat{j} + (2C(2) - 4) \hat{k} \]
This simplifies to:
\[ \vec{v} = A \hat{i} + (12B - 2) \hat{j} + (4C - 4) \hat{k} \]
Setting this equal to \( 3 \hat{i} + 22 \hat{j} \):
- From the \( \hat{i} \) component: \( A = 3 \)
- From the \( \hat{j} \) component: \( 12B - 2 = 22 \) \( \Rightarrow 12B = 24 \) \( \Rightarrow B = 2 \)
- From the \( \hat{k} \) component: \( 4C - 4 = 0 \) \( \Rightarrow 4C = 4 \) \( \Rightarrow C = 1 \)

4. **Substitute \( A, B, C \) back into the velocity equation**: Now we have \( A = 3 \), \( B = 2 \), and \( C = 1 \). The velocity vector becomes:
\[ \vec{v} = 3 \hat{i} + (3(2)(t^2) - 2) \hat{j} + (2(1)t - 4) \hat{k} \]
Simplifying this gives:
\[ \vec{v} = 3 \hat{i} + (6t^2 - 2) \hat{j} + (2t - 4) \hat{k} \]

5. **Calculate the velocity at \( t = 4 \) seconds**:
\[ \vec{v} = 3 \hat{i} + (6(4^2) - 2) \hat{j} + (2(4) - 4) \hat{k} \]
Calculating each component:
- \( 4^2 = 16 \) so \( 6(16) - 2 = 96 - 2 = 94 \)
- \( 2(4) - 4 = 8 - 4 = 4 \)

Thus, the velocity vector at \( t = 4 \) seconds is:
\[ \vec{v} = 3 \hat{i} + 94 \hat{j} + 4 \hat{k} \]

### Final Answer:
\[ \vec{v} = 3 \hat{i} + 94 \hat{j} + 4 \hat{k} \text{ m/s} \]