A particle is projected vertically upwards with velocity `40m//s.` Find the displacement and distance travelled by the particle in
(a) `2s` (b) `4s` (c) `6s` Take `g=10 m//s^2`

Explanation

To solve the problem, we will calculate the displacement and distance traveled by a particle projected vertically upwards with an initial velocity of \( u = 40 \, \text{m/s} \) under the influence of gravity \( g = 10 \, \text{m/s}^2 \).

### Given:
- Initial velocity, \( u = 40 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (acting downwards)
- Time intervals: \( t = 2 \, \text{s}, 4 \, \text{s}, 6 \, \text{s} \)

### Formulas:
1. Displacement:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( a = -g \) (since gravity acts downwards).

2. Distance:
- For the first 4 seconds, distance equals displacement.
- For time greater than 4 seconds, we need to calculate the distance separately.

### Step-by-Step Solution:

#### (a) For \( t = 2 \, \text{s} \):
1. Calculate displacement:
\[ s = ut + \frac{1}{2}(-g)t^2 = 40 \times 2 + \frac{1}{2}(-10)(2^2) \]
\[ s = 80 - 20 = 60 \, \text{m} \]

2. Since the particle is still moving upwards, the distance traveled is also:
\[ \text{Distance} = 60 \, \text{m} \]

#### (b) For \( t = 4 \, \text{s} \):
1. Calculate displacement:
\[ s = ut + \frac{1}{2}(-g)t^2 = 40 \times 4 + \frac{1}{2}(-10)(4^2) \]
\[ s = 160 - 80 = 80 \, \text{m} \]

2. Again, the particle is still moving upwards, so:
\[ \text{Distance} = 80 \, \text{m} \]

#### (c) For \( t = 6 \, \text{s} \):
1. Calculate displacement:
\[ s = ut + \frac{1}{2}(-g)t^2 = 40 \times 6 + \frac{1}{2}(-10)(6^2) \]
\[ s = 240 - 180 = 60 \, \text{m} \]

2. For distance, we need to consider the motion:
- The particle goes up for 4 seconds and then comes down for 2 seconds.
- Distance traveled upwards in 4 seconds = 80 m (as calculated).
- Distance traveled downwards in 2 seconds:
- Initial velocity when coming down = 0 (at the peak).
- Distance down in 2 seconds:
\[ s_{down} = \frac{1}{2} g t^2 = \frac{1}{2} \times 10 \times (2^2) = 20 \, \text{m} \]
- Total distance:
\[ \text{Total Distance} = 80 + 20 = 100 \, \text{m} \]

### Summary of Results:
- At \( t = 2 \, \text{s} \): Displacement = 60 m, Distance = 60 m
- At \( t = 4 \, \text{s} \): Displacement = 80 m, Distance = 80 m
- At \( t = 6 \, \text{s} \): Displacement = 60 m, Distance = 100 m