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:
  \(S_{t+2} = 100 + \frac{50}{2}(3)\)
 \(S_{t+2} = 100 + 25 \times 3\)
 \(S_{t+2} = 100 + 75 = 175 \, \text{m}\)

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

Answer 2

The problem asks for the distance travelled by a particle in the \((t+2)^{\text{th}}\) second, given its displacement and the increase in its velocity during the time interval from \(t\) to \((t+1)\) seconds.

Concept Used:

The problem can be solved using the equations of motion for a particle undergoing uniformly accelerated motion. The key steps are:

1. Determine the acceleration of the particle from the given increase in velocity over a specific time interval. The formula for constant acceleration is \( a = \frac{\Delta v}{\Delta t} \).
2. Use the given displacement and the calculated acceleration to find the velocity of the particle at the beginning of the first time interval (at time \(t\)). The relevant equation of motion is \( s = ut + \frac{1}{2}at^2 \).
3. Calculate the velocity at the beginning of the next time interval (at time \(t+1\)). The relevant equation is \( v = u + at \).
4. Finally, calculate the distance travelled during the \((t+2)^{\text{th}}\) second using the velocity at time \(t+1\) as the initial velocity for this interval.

Step-by-Step Solution:

Step 1: Calculate the acceleration of the particle.

The time interval is from \(t\) to \((t+1)\), so the duration is \( \Delta t = (t+1) - t = 1 \) s.

The increase in velocity during this interval is \( \Delta v = 50 \) m/s.

Assuming the acceleration is constant, it is given by:

\[ a = \frac{\Delta v}{\Delta t} = \frac{50 \, \text{m/s}}{1 \, \text{s}} = 50 \, \text{m/s}^2 \]

Step 2: Calculate the velocity of the particle at time \(t\).

Let the velocity of the particle at time \(t\) be \(u\). The displacement in the time interval from \(t\) to \((t+1)\) is given as 125 m. Using the equation of motion \( s = u(\Delta t) + \frac{1}{2}a(\Delta t)^2 \):

\[ 125 = u(1) + \frac{1}{2}(50)(1)^2 \] \[ 125 = u + 25 \]

Solving for \(u\):

\[ u = 125 - 25 = 100 \, \text{m/s} \]

Step 3: Calculate the velocity of the particle at time \((t+1)\).

The distance travelled in the \((t+2)^{\text{th}}\) second corresponds to the displacement in the time interval from \((t+1)\) to \((t+2)\). We need the velocity at the beginning of this interval, i.e., at time \((t+1)\). Let this velocity be \(u'\).

\[ u' = u + a(\Delta t) = 100 \, \text{m/s} + (50 \, \text{m/s}^2)(1 \, \text{s}) = 150 \, \text{m/s} \]

Final Computation & Result:

Step 4: Calculate the distance travelled in the \((t+2)^{\text{th}}\) second.

The time interval is from \((t+1)\) to \((t+2)\), so the duration is \( \Delta t = 1 \) s. The initial velocity for this interval is \( u' = 150 \) m/s.

The distance travelled (\(s'\)) in this second is:

\[ s' = u'(\Delta t) + \frac{1}{2}a(\Delta t)^2 \] \[ s' = (150)(1) + \frac{1}{2}(50)(1)^2 \] \[ s' = 150 + 25 = 175 \, \text{m} \]

Since the initial velocity is positive and the acceleration is positive, the particle does not change direction, so the distance travelled is equal to the displacement.

The distance travelled by the particle in the \((t+2)^{\text{th}}\) second is 175 m.