Question

The displacement and the increase in the velocity of a moving particle in the time interval of \( t \) to \( (t + 1) \) s are 125 m and 50 m/s, respectively. The distance travelled by the particle in \( (t + 2)^{\text{th}} \) s is \( \_\_\_\_ \) m.

Answer 1

Correct Answer: 175

Given Information:
 - Displacement from \( t \) to \( t+1 \): \( \Delta x = 125 \, \text{m} \)
 - Velocity increase from \( t \) to \( t+1 \): \( \Delta v = 50 \, \text{m/s} \)

Now, Calculate Acceleration:
 Using the formula:
 \(a = \frac{\Delta v}{\Delta t}\)
  Substituting \( \Delta v = 50 \, \text{m/s} \) and \( \Delta t = 1 \, \text{s} \):
 \(a = \frac{50}{1} = 50 \, \text{m/s}^2\)

Equation of Motion:
 The formula for distance traveled in the \( n \)-th second is:
\(S_n = u + \frac{a}{2} (2n - 1)\)

Find Initial Velocity (\( u \)):
Distance traveled in the \( (t+1) \)-th second is \( S_{t+1} = 125 \, \text{m} \). Substituting \( n = 1 \), \( a = 50 \, \text{m/s}^2 \), and \( S_{t+1} = 125 \):
 \(125 = u + \frac{50}{2}(2 \cdot 1 - 1)\)
  Simplifying:
  \(125 = u + 25\)
  \(u = 125 - 25 = 100 \, \text{m/s}\)

Now, Calculate Distance for \( (t+2) \)-th Second:
 Substituting \( u = 100 \, \text{m/s} \), \( a = 50 \, \text{m/s}^2 \), and \( n = 2 \) into the equation:

\(S_{t+2} = u + \frac{a}{2}(2 \cdot 2 - 1)\)
  Simplifying:
  \(S_{t+2} = 100 + \frac{50}{2}(3)\)
 \(S_{t+2} = 100 + 25 \times 3\)
 \(S_{t+2} = 100 + 75 = 175 \, \text{m}\)

\(\boxed{\text{The distance traveled in the } (t+2)\text{-th second is } 175 \, \text{m}.}\)