If the velocity v of a particle moving along a straight line decreases linearly with its displacement from `20 m//s` to a value approaching zero at `s = 30 m,` determine the acceleration of the particle when `s = 15 m.`

Explanation

To solve the problem, we need to determine the acceleration of the particle when its displacement \( s = 15 \, m \). We know that the velocity \( v \) decreases linearly from \( 20 \, m/s \) at \( s = 0 \, m \) to \( 0 \, m/s \) at \( s = 30 \, m \).

### Step-by-Step Solution:

1. **Establish the relationship between velocity and displacement**:
Since the velocity decreases linearly from \( 20 \, m/s \) to \( 0 \, m/s \) as displacement increases from \( 0 \, m \) to \( 30 \, m \), we can express this relationship as:
\[ v = 20 - \frac{20}{30} s \]
Simplifying this gives:
\[ v = 20 - \frac{2}{3}s \]

2. **Find the velocity at \( s = 15 \, m \)**:
Substitute \( s = 15 \, m \) into the velocity equation:
\[ v = 20 - \frac{2}{3} \cdot 15 \]
\[ v = 20 - 10 = 10 \, m/s \]

3. **Use the kinematic relation to find acceleration**:
We know that acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} \]
We can also use the chain rule to express acceleration in terms of displacement:
\[ a = \frac{dv}{ds} \cdot \frac{ds}{dt} = \frac{dv}{ds} \cdot v \]

4. **Calculate \( \frac{dv}{ds} \)**:
Differentiate the velocity equation with respect to \( s \):
\[ \frac{dv}{ds} = -\frac{2}{3} \]

5. **Substitute \( v \) and \( \frac{dv}{ds} \) into the acceleration equation**:
Now, substituting \( v = 10 \, m/s \) and \( \frac{dv}{ds} = -\frac{2}{3} \):
\[ a = \left(-\frac{2}{3}\right) \cdot 10 \]
\[ a = -\frac{20}{3} \, m/s^2 \]

6. **Final Result**:
The acceleration of the particle when \( s = 15 \, m \) is:
\[ a = -\frac{20}{3} \, m/s^2 \approx -6.67 \, m/s^2 \]