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 a particle whose velocity decreases linearly with displacement. Here’s a step-by-step solution:

### Step 1: Establish the relationship between velocity and displacement
The problem states that the velocity \( v \) decreases linearly from \( 20 \, \text{m/s} \) at \( s = 0 \) to \( 0 \, \text{m/s} \) at \( s = 30 \, \text{m} \). We can express this relationship as a linear equation:
\[ v = mx + c \]
where \( m \) is the slope and \( c \) is the y-intercept.

### Step 2: Determine the slope (m) and y-intercept (c)
From the given points:
- At \( s = 0 \), \( v = 20 \, \text{m/s} \) (this gives us \( c = 20 \))
- At \( s = 30 \), \( v = 0 \, \text{m/s} \)

Using the two points to find the slope:
\[ m = \frac{v_2 - v_1}{s_2 - s_1} = \frac{0 - 20}{30 - 0} = \frac{-20}{30} = -\frac{2}{3} \]
Thus, the equation for velocity as a function of displacement \( s \) becomes:
\[ v = -\frac{2}{3}s + 20 \]

### Step 3: Find the velocity at \( s = 15 \, \text{m} \)
Now, we can calculate the velocity when \( s = 15 \, \text{m} \):
\[ v = -\frac{2}{3}(15) + 20 = -10 + 20 = 10 \, \text{m/s} \]

### Step 4: Find the derivative of velocity with respect to displacement
Next, we need to find \( \frac{dv}{ds} \):
\[ \frac{dv}{ds} = -\frac{2}{3} \, \text{(constant)} \]

### Step 5: Calculate the acceleration using the formula
The acceleration \( a \) can be expressed in terms of velocity and its derivative:
\[ a = v \frac{dv}{ds} \]
Substituting the values we have:
\[ a = 10 \left(-\frac{2}{3}\right) = -\frac{20}{3} \, \text{m/s}^2 \]

### Final Answer
Thus, the acceleration of the particle when \( s = 15 \, \text{m} \) is:
\[ a = -\frac{20}{3} \, \text{m/s}^2 \]

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