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 Data:
- Displacement over a time interval of \( \Delta t = 1 \, \text{s} \): \( \Delta x = 125 \, \text{m} \)
 - Change in velocity over the same interval \( \Delta t = 1 \, \text{s} \): \( \Delta v = 50 \, \text{m/s} \)

Acceleration Calculation:
 Using the definition of acceleration:
 \(a = \frac{\Delta v}{\Delta t}\)
  Substituting the given values \( \Delta v = 50 \, \text{m/s} \) and \( \Delta t = 1 \, \text{s} \):
 \(a = \frac{50 \, \text{m/s}}{1 \, \text{s}} = 50 \, \text{m/s}^2\)

Equation for Distance in the n-th Second:
 The formula relating distance traveled in the \( n \)-th second (\( S_n \)), initial velocity (\( u \)), and acceleration (\( a \)) is:
\(S_n = u + \frac{a}{2} (2n - 1)\)

Determine Initial Velocity (\( u \)):
The displacement in the \( (t+1) \)-th second is \( S_{t+1} = 125 \, \text{m} \). Using the formula with \( n = 1 \), \( a = 50 \, \text{m/s}^2 \), and \( S_{t+1} = 125 \, \text{m} \):
 \(125 = u + \frac{50}{2}(2 \cdot 1 - 1)\)
  Simplification yields:
  \(125 = u + 25(1)\)
  \(125 = u + 25\)
  \(u = 125 - 25 = 100 \, \text{m/s}\)

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

\(S_{t+2} = 100 + \frac{50}{2}(2 \cdot 2 - 1)\)
  Simplification:
  \(S_{t+2} = 100 + 25(4 - 1)\)
  \(S_{t+2} = 100 + 25(3)\)
  \(S_{t+2} = 100 + 75\)
  \(S_{t+2} = 175 \, \text{m}\)

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