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.

Options

A. `10`
B. `26`
C. `13`
D. `20`

Explanation

To solve the problem, we will break it down into steps to find the distance traveled by the particle in 5 seconds after it passes point A.

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

### Step 2: Calculate the time taken to come to rest
We need to find out how long it takes for the particle to come to a stop (final velocity, v = 0).

Using the equation of motion:
\[ v = u + at \]

Substituting the known values:
\[ 0 = 8 + (-4)t \]
\[ 0 = 8 - 4t \]
\[ 4t = 8 \]
\[ t = 2 \text{ seconds} \]

### Step 3: Calculate the distance traveled in the first 2 seconds
Now we will calculate the distance traveled in the first 2 seconds until the particle comes to rest.

Using the equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]

Substituting the known values:
\[ s = 8(2) + \frac{1}{2}(-4)(2^2) \]
\[ s = 16 - \frac{1}{2}(4)(4) \]
\[ s = 16 - 8 \]
\[ s = 8 \text{ meters} \]

### Step 4: Calculate the distance traveled after coming to rest
After 2 seconds, the particle comes to rest. It will then start moving in the opposite direction due to negative acceleration. We need to calculate the distance traveled in the next 3 seconds (from t = 2s to t = 5s).

Now, the initial velocity for this phase is 0 m/s (as it has come to rest), and the acceleration is now +4 m/s² (in the opposite direction).

Using the same equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]

Substituting the known values (u = 0, a = 4 m/s², t = 3 seconds):
\[ s = 0(3) + \frac{1}{2}(4)(3^2) \]
\[ s = 0 + \frac{1}{2}(4)(9) \]
\[ s = 0 + 18 \]
\[ s = 18 \text{ meters} \]

### Step 5: Calculate the total distance traveled
Now, we will add the distances from both phases:
- Distance traveled in the first 2 seconds = 8 meters
- Distance traveled in the next 3 seconds = 18 meters

Total distance:
\[ \text{Total distance} = 8 + 18 = 26 \text{ meters} \]

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

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