A particle is projected vertically upwards with an initial velocity of `40 m//s.` Find the displacement and distance covered by the particle in `6 s.` Take `g= 10 m//s^2.`

Explanation

To solve the problem of a particle projected vertically upwards with an initial velocity of 40 m/s, we will calculate both the displacement and the distance covered by the particle in 6 seconds, taking \( g = 10 \, \text{m/s}^2 \).

### Step 1: Calculate Displacement
The formula for displacement \( s \) when an object is under uniform acceleration is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( u \) = initial velocity = 40 m/s
- \( a \) = acceleration = -g = -10 m/s² (negative because it is acting downwards)
- \( t \) = time = 6 s

Substituting the values into the formula:
\[ s = 40 \times 6 + \frac{1}{2} \times (-10) \times (6^2) \]
Calculating each term:
\[ s = 240 + \frac{1}{2} \times (-10) \times 36 \]
\[ s = 240 - 180 \]
\[ s = 60 \, \text{m} \]

### Step 2: Calculate Distance
To find the total distance traveled, we need to consider the motion of the particle. The particle will rise to its highest point and then fall back down.

#### Step 2.1: Find the time to reach the highest point
At the highest point, the final velocity \( v = 0 \). We can use the equation:
\[ v = u + at \]
Setting \( v = 0 \):
\[ 0 = 40 - 10t \]
Solving for \( t \):
\[ 10t = 40 \implies t = 4 \, \text{s} \]

#### Step 2.2: Calculate distance to the highest point
Using the same displacement formula for the time of 4 seconds:
\[ s_1 = ut + \frac{1}{2} a t^2 \]
Substituting \( t = 4 \):
\[ s_1 = 40 \times 4 + \frac{1}{2} \times (-10) \times (4^2) \]
Calculating:
\[ s_1 = 160 - 80 \]
\[ s_1 = 80 \, \text{m} \]

#### Step 2.3: Calculate distance from the highest point to 6 seconds
The time taken to fall back down after reaching the highest point is:
\[ t_2 = 6 - 4 = 2 \, \text{s} \]
The distance fallen during this time can be calculated using:
\[ s_2 = ut + \frac{1}{2} a t^2 \]
Here, the initial velocity \( u = 0 \) (at the highest point):
\[ s_2 = 0 \times 2 + \frac{1}{2} \times 10 \times (2^2) \]
Calculating:
\[ s_2 = 0 + \frac{1}{2} \times 10 \times 4 \]
\[ s_2 = 20 \, \text{m} \]

#### Step 2.4: Total distance covered
The total distance \( d \) is the sum of the distance to the highest point and the distance fallen:
\[ d = s_1 + s_2 = 80 + 20 = 100 \, \text{m} \]

### Final Results
- **Displacement**: \( 60 \, \text{m} \)
- **Distance**: \( 100 \, \text{m} \)