The momentum p of a particle changes with the t according to the relation `dp/dt=(10N)+(2N/s)t`. If the momentum is zero at t=0, what will the momentum be at t=10?.

Explanation

To solve the problem, we need to find the momentum \( p \) of a particle at \( t = 10 \) seconds given the rate of change of momentum \( \frac{dp}{dt} = 10 + 2t \) and the initial condition that \( p(0) = 0 \).

### Step-by-Step Solution:

1. **Write down the given equation:**
\[ \frac{dp}{dt} = 10 + 2t \]

2. **Integrate both sides with respect to time \( t \):**
To find \( p \), we need to integrate \( \frac{dp}{dt} \):
\[ dp = (10 + 2t) dt \]

3. **Integrate the right-hand side:**
\[ \int dp = \int (10 + 2t) dt \]
The integral of \( dp \) is simply \( p \). Now we integrate the right-hand side:
\[ p = \int 10 dt + \int 2t dt \]
\[ p = 10t + t^2 + C \]
where \( C \) is the constant of integration.

4. **Use the initial condition to find \( C \):**
We know that \( p(0) = 0 \):
\[ p(0) = 10(0) + (0)^2 + C = 0 \]
This implies that \( C = 0 \).

5. **Substitute \( C \) back into the equation:**
\[ p = 10t + t^2 \]

6. **Calculate \( p \) at \( t = 10 \):**
\[ p(10) = 10(10) + (10)^2 \]
\[ p(10) = 100 + 100 = 200 \]

7. **Final answer:**
The momentum at \( t = 10 \) seconds is:
\[ p(10) = 200 \, \text{N·s} \]