The initial velocity of the particle is `10m//sec` and its retardation is `2m//sec^(2)`. The distance moved by the particle in 5th second of its motion is

Options

A. 1m
B. 19m
C. 50m
D. 75m

Answer

To find the distance moved by the particle in the 5th second of its motion, we can use the equations of motion.

### Step 1: Understand the given values
- Initial velocity (u) = 10 m/s
- Retardation (a) = -2 m/s² (since it is in the opposite direction of the velocity)

### Step 2: Calculate the velocity at the end of the 5th second
We can use the formula for velocity:
\[ v = u + at \]
Where:
- \( v \) = final velocity after time \( t \)
- \( u \) = initial velocity
- \( a \) = acceleration (retardation in this case)
- \( t \) = time in seconds

For the 5th second, \( t = 5 \):
\[ v = 10 + (-2) \cdot 5 \]
\[ v = 10 - 10 \]
\[ v = 0 \, \text{m/s} \]

### Step 3: Calculate the distance moved in the first 5 seconds
We can use the formula for distance:
\[ s = ut + \frac{1}{2} a t^2 \]

Substituting the values:
\[ s = 10 \cdot 5 + \frac{1}{2} \cdot (-2) \cdot (5^2) \]
\[ s = 50 - \frac{1}{2} \cdot 2 \cdot 25 \]
\[ s = 50 - 25 \]
\[ s = 25 \, \text{m} \]

### Step 4: Calculate the distance moved in the first 4 seconds
Now we need to find the distance moved in the first 4 seconds to find the distance moved in the 5th second:
\[ s_4 = u \cdot 4 + \frac{1}{2} a \cdot (4^2) \]
\[ s_4 = 10 \cdot 4 + \frac{1}{2} \cdot (-2) \cdot (4^2) \]
\[ s_4 = 40 - \frac{1}{2} \cdot 2 \cdot 16 \]
\[ s_4 = 40 - 16 \]
\[ s_4 = 24 \, \text{m} \]

### Step 5: Calculate the distance moved in the 5th second
The distance moved in the 5th second is given by:
\[ s_5 = s_5 - s_4 \]
\[ s_5 = 25 - 24 \]
\[ s_5 = 1 \, \text{m} \]

### Final Answer
The distance moved by the particle in the 5th second of its motion is **1 meter**.

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