A body starts from rest from the origin with an acceleration of `6m//s^(2)` along the x-axis and `8m//s^(2)` along the y-axis. Its distance from the origin after 4 seconds will be

Options

A. 56 m
B. 64m
C. 80 m
D. 128m

Answer

To solve the problem step-by-step, we need to calculate the distance traveled by the body in both the x and y directions after 4 seconds, and then use the Pythagorean theorem to find the total distance from the origin.

### Step 1: Identify the given values
- Initial velocity (u) = 0 m/s (since the body starts from rest)
- Acceleration in the x-direction (a_x) = 6 m/s²
- Acceleration in the y-direction (a_y) = 8 m/s²
- Time (t) = 4 seconds

### Step 2: Calculate the distance traveled in the x-direction (s_x)
Using the formula for distance under constant acceleration:
\[ s_x = ut + \frac{1}{2} a_x t^2 \]
Since the initial velocity (u) is 0, the equation simplifies to:
\[ s_x = \frac{1}{2} a_x t^2 \]
Substituting the values:
\[ s_x = \frac{1}{2} \times 6 \, \text{m/s}^2 \times (4 \, \text{s})^2 \]
\[ s_x = \frac{1}{2} \times 6 \times 16 \]
\[ s_x = 48 \, \text{m} \]

### Step 3: Calculate the distance traveled in the y-direction (s_y)
Using the same formula:
\[ s_y = ut + \frac{1}{2} a_y t^2 \]
Again, since the initial velocity (u) is 0:
\[ s_y = \frac{1}{2} a_y t^2 \]
Substituting the values:
\[ s_y = \frac{1}{2} \times 8 \, \text{m/s}^2 \times (4 \, \text{s})^2 \]
\[ s_y = \frac{1}{2} \times 8 \times 16 \]
\[ s_y = 64 \, \text{m} \]

### Step 4: Calculate the total distance from the origin (r)
Using the Pythagorean theorem:
\[ r = \sqrt{s_x^2 + s_y^2} \]
Substituting the values:
\[ r = \sqrt{(48 \, \text{m})^2 + (64 \, \text{m})^2} \]
\[ r = \sqrt{2304 + 4096} \]
\[ r = \sqrt{6400} \]
\[ r = 80 \, \text{m} \]

### Final Answer
The distance from the origin after 4 seconds is **80 meters**.

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