Velocity of particle moving along positive x-direction is `v = (40-10t)m//s`. Here,t is in seconds. At time `t=0,` tha x coordinate of particle is zero. Find the time when the particle is at a distance of 60 m from origin.

Explanation

To solve the problem, we need to determine the time when a particle, moving along the positive x-direction with a velocity given by \( v = 40 - 10t \) m/s, is at a distance of 60 m from the origin. The initial position of the particle at \( t = 0 \) is \( x = 0 \).

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The velocity of the particle is given by \( v = 40 - 10t \) m/s.
- The initial position \( x(0) = 0 \) m.
- We need to find the time \( t \) when the particle is at \( x = 60 \) m.

2. **Identify Initial Conditions**:
- The initial velocity \( u = 40 \) m/s (at \( t = 0 \)).
- The acceleration \( a \) can be derived from the velocity equation:
\[ a = \frac{dv}{dt} = -10 \text{ m/s}^2 \]

3. **Use the Equation of Motion**:
- We can use the second equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
- Here, \( S = 60 \) m, \( u = 40 \) m/s, and \( a = -10 \) m/s².

4. **Substitute Values into the Equation**:
\[ 60 = 40t + \frac{1}{2}(-10)t^2 \]
Simplifying this gives:
\[ 60 = 40t - 5t^2 \]
Rearranging the equation:
\[ 5t^2 - 40t + 60 = 0 \]
Dividing through by 5:
\[ t^2 - 8t + 12 = 0 \]

5. **Solve the Quadratic Equation**:
- Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1, b = -8, c = 12 \):
\[ t = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} \]
\[ t = \frac{8 \pm \sqrt{64 - 48}}{2} \]
\[ t = \frac{8 \pm \sqrt{16}}{2} \]
\[ t = \frac{8 \pm 4}{2} \]
This gives us two possible solutions:
\[ t_1 = \frac{12}{2} = 6 \text{ seconds} \]
\[ t_2 = \frac{4}{2} = 2 \text{ seconds} \]

6. **Conclusion**:
- The particle is at a distance of 60 m from the origin at two different times: \( t = 2 \) seconds and \( t = 6 \) seconds.