A particle movig with a uniformacceleration travels ` 24 ` metre and ` 64` metre in first two successive intervals of ` 4` seconds each. Its initial velocity is.

Options

A. (a) ` 1 m//s`
B. (b) 2 m//s`
C. (c ) ` 5 m//s`
D. (d)` 10 m//s`.

Explanation

To find the initial velocity of a particle moving with uniform acceleration, we can use the equations of motion. Here’s a step-by-step solution:

### Given:
1. Distance covered in the first 4 seconds, \( s_1 = 24 \) m
2. Distance covered in the next 4 seconds, \( s_2 = 64 \) m
3. Time intervals, \( t_1 = 4 \) s and \( t_2 = 4 \) s

### Step 1: Write the equations of motion for the first interval
For the first 4 seconds, we use the equation of motion:
\[ s_1 = ut_1 + \frac{1}{2} a t_1^2 \]
Substituting the known values:
\[ 24 = u(4) + \frac{1}{2} a (4^2) \]
This simplifies to:
\[ 24 = 4u + 8a \quad \text{(Equation 1)} \]

### Step 2: Write the equations of motion for the second interval
For the next 4 seconds, the initial velocity is now \( v = u + at_1 \). We can express the second equation as:
\[ s_2 = vt_2 + \frac{1}{2} a t_2^2 \]
Substituting the known values:
\[ 64 = (u + 4a)(4) + \frac{1}{2} a (4^2) \]
This simplifies to:
\[ 64 = 4u + 16a + 8a \]
Combining like terms gives us:
\[ 64 = 4u + 24a \quad \text{(Equation 2)} \]

### Step 3: Solve the equations simultaneously
We now have two equations:
1. \( 4u + 8a = 24 \) (Equation 1)
2. \( 4u + 24a = 64 \) (Equation 2)

Subtract Equation 1 from Equation 2:
\[ (4u + 24a) - (4u + 8a) = 64 - 24 \]
This simplifies to:
\[ 16a = 40 \]
Thus,
\[ a = \frac{40}{16} = 2.5 \, \text{m/s}^2 \]

### Step 4: Substitute \( a \) back to find \( u \)
Now we can substitute \( a \) back into Equation 1:
\[ 4u + 8(2.5) = 24 \]
This simplifies to:
\[ 4u + 20 = 24 \]
Thus,
\[ 4u = 4 \quad \Rightarrow \quad u = 1 \, \text{m/s} \]

### Final Answer:
The initial velocity \( u \) of the particle is \( 1 \, \text{m/s} \).

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