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

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

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