A particle is moving in a straight line. All the physical quantities are to be measured in MKS system. Square of the magnitude of its instantaneous velocity is found to be ten times its instantaneous displacement . What is the acceleration of the particle?

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the given information
We are given that the square of the magnitude of the instantaneous velocity (v²) is equal to ten times the instantaneous displacement (s). Mathematically, this can be expressed as:
\[ v^2 = 10s \]

### Step 2: Identify the relevant equations
In the case of uniform acceleration, we can use the kinematic equation that relates velocity, displacement, and acceleration:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration
- \( s \) = displacement

### Step 3: Set initial conditions
Since the problem does not mention any initial velocity, we can assume that the particle starts from rest. Therefore:
\[ u = 0 \]

### Step 4: Substitute into the kinematic equation
Substituting \( u = 0 \) into the kinematic equation gives:
\[ v^2 = 0 + 2as \]
or simply:
\[ v^2 = 2as \]

### Step 5: Equate the two expressions for v²
From Step 1, we have \( v^2 = 10s \). From Step 4, we have \( v^2 = 2as \). Setting these two equations equal to each other:
\[ 10s = 2as \]

### Step 6: Simplify the equation
We can divide both sides by \( s \) (assuming \( s \neq 0 \)):
\[ 10 = 2a \]

### Step 7: Solve for acceleration (a)
Now, we can solve for \( a \):
\[ a = \frac{10}{2} = 5 \, \text{m/s}^2 \]

### Conclusion
The acceleration of the particle is \( 5 \, \text{m/s}^2 \).

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