The velocity of an object at t =0 is ` vecv_(0) =- 4 hatj ` m/s . It moves in plane with constant acceleration ` veca = ( 3hati + 8 hatj) m//s^(2)`. What is its velocity after 1 s?

Answer

To find the velocity of the object after 1 second, we can use the equation of motion in vector form:

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- Initial velocity \( \vec{v}_0 = -4 \hat{j} \) m/s
- Acceleration \( \vec{a} = 3 \hat{i} + 8 \hat{j} \) m/s²
- Time \( t = 1 \) s

2. **Write the equation of motion:**
The equation for the final velocity \( \vec{v} \) is given by:
\[ \vec{v} = \vec{v}_0 + \vec{a}t \]

3. **Substitute the known values:**
\[ \vec{v} = (-4 \hat{j}) + (3 \hat{i} + 8 \hat{j}) \cdot 1 \]
This simplifies to:
\[ \vec{v} = -4 \hat{j} + 3 \hat{i} + 8 \hat{j} \]

4. **Combine the components:**
Combine the \( \hat{j} \) components:
\[ \vec{v} = 3 \hat{i} + (-4 + 8) \hat{j} = 3 \hat{i} + 4 \hat{j} \]

5. **Calculate the magnitude of the velocity:**
The magnitude of the velocity \( |\vec{v}| \) can be calculated using the Pythagorean theorem:
\[ |\vec{v}| = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m/s} \]

6. **Final Result:**
The velocity of the object after 1 second is:
\[ \vec{v} = 3 \hat{i} + 4 \hat{j} \text{ m/s} \quad \text{and its magnitude is } 5 \text{ m/s} \]