A particle moving along a straight line with a constant acceleration of `-4 m//s^2` passes through a point `A` on the line with a velocity of `+ 8 m//s` at some moment. Find the distance travelled by the particle in `5` seconds after that moment.

Explanation

To solve the problem step by step, we will use the equations of motion under constant acceleration.

### Step 1: Identify the given values
- Initial velocity (u) = +8 m/s (at point A)
- Acceleration (a) = -4 m/s²
- Time (t) = 5 seconds

### Step 2: Calculate the time when the velocity becomes zero
Using the formula for final velocity:
\[ v = u + at \]
Setting \( v = 0 \) (the point when the particle stops):
\[ 0 = 8 - 4t \]
Rearranging gives:
\[ 4t = 8 \]
\[ t = 2 \text{ seconds} \]

### Step 3: Calculate the distance traveled in the first 2 seconds
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values for the first 2 seconds:
\[ s_1 = 8 \cdot 2 + \frac{1}{2} \cdot (-4) \cdot (2^2) \]
Calculating:
\[ s_1 = 16 - \frac{1}{2} \cdot 4 \cdot 4 \]
\[ s_1 = 16 - 8 \]
\[ s_1 = 8 \text{ meters} \]

### Step 4: Calculate the distance traveled in the next 3 seconds (from 2 to 5 seconds)
At \( t = 2 \) seconds, the velocity is 0 m/s. Now, we will calculate the distance for the next 3 seconds:
Using the same equation of motion:
\[ s_2 = u t + \frac{1}{2} a t^2 \]
Here, the initial velocity \( u = 0 \) (since it has just stopped):
\[ s_2 = 0 \cdot 3 + \frac{1}{2} \cdot (-4) \cdot (3^2) \]
Calculating:
\[ s_2 = 0 - \frac{1}{2} \cdot 4 \cdot 9 \]
\[ s_2 = -18 \text{ meters} \]

### Step 5: Calculate the total distance traveled
The total distance traveled is the sum of the absolute values of \( s_1 \) and \( s_2 \):
\[ \text{Total distance} = |s_1| + |s_2| \]
\[ \text{Total distance} = 8 + 18 = 26 \text{ meters} \]

### Final Answer
The distance traveled by the particle in 5 seconds after passing point A is **26 meters**.

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