The speed of a body moving on a straight trach varies according to `v=(4)/(5)t^2+3` for `0lttlt5` sec and `v=3t+8` for `tgt5` sec. If dinstances are in meters, find the distance moved by a particle at the end of 10 sec.

Explanation

To solve the problem of finding the distance moved by a particle at the end of 10 seconds, we will break the process down into steps.

### Step-by-Step Solution:

1. **Identify the velocity functions**:
- For \(0 < t < 5\) seconds:
\[ v = \frac{4}{5}t^2 + 3 \]
- For \(t \geq 5\) seconds:
\[ v = 3t + 8 \]

2. **Calculate the distance for the first interval (0 to 5 seconds)**:
- The distance \(x_1\) can be found by integrating the velocity function from \(t = 0\) to \(t = 5\):
\[ x_1 = \int_0^5 v \, dt = \int_0^5 \left(\frac{4}{5}t^2 + 3\right) dt \]

3. **Perform the integration**:
- The integral can be split:
\[ x_1 = \int_0^5 \frac{4}{5}t^2 \, dt + \int_0^5 3 \, dt \]
- Calculate each part:
- For \(\int_0^5 \frac{4}{5}t^2 \, dt\):
\[ = \frac{4}{5} \cdot \left[\frac{t^3}{3}\right]_0^5 = \frac{4}{5} \cdot \frac{125}{3} = \frac{100}{3} \]
- For \(\int_0^5 3 \, dt\):
\[ = 3[t]_0^5 = 3 \cdot 5 = 15 \]
- Combine the results:
\[ x_1 = \frac{100}{3} + 15 = \frac{100}{3} + \frac{45}{3} = \frac{145}{3} \text{ meters} \]

4. **Calculate the distance for the second interval (5 to 10 seconds)**:
- The distance \(x_2\) can be found by integrating the second velocity function from \(t = 5\) to \(t = 10\):
\[ x_2 = \int_5^{10} (3t + 8) \, dt \]

5. **Perform the integration**:
- Calculate the integral:
\[ x_2 = \int_5^{10} (3t + 8) \, dt = \left[\frac{3t^2}{2} + 8t\right]_5^{10} \]
- Evaluate at the limits:
- For \(t = 10\):
\[ = \frac{3(10^2)}{2} + 8(10) = \frac{300}{2} + 80 = 150 + 80 = 230 \]
- For \(t = 5\):
\[ = \frac{3(5^2)}{2} + 8(5) = \frac{75}{2} + 40 = 37.5 + 40 = 77.5 \]
- Combine the results:
\[ x_2 = 230 - 77.5 = 152.5 \text{ meters} \]

6. **Total distance moved in 10 seconds**:
- Combine the distances from both intervals:
\[ x_{\text{total}} = x_1 + x_2 = \frac{145}{3} + 152.5 \]
- Convert \(152.5\) to a fraction:
\[ 152.5 = \frac{305}{2} \]
- Find a common denominator (6):
\[ x_{\text{total}} = \frac{145}{3} + \frac{915}{6} = \frac{290}{6} + \frac{915}{6} = \frac{1205}{6} \approx 200.83 \text{ meters} \]

### Final Answer:
The total distance moved by the particle at the end of 10 seconds is approximately **200.83 meters**.