A particle moves clockwise in a circle of radius `1 m` with centre at `(x, y) = (1m, 0)`. It startsx at rest at the origin at time `t = 0`. Its speed increase at the constant rate of `((pi)/(2)) bm//s^(2)`. If the net acceleration at `t = 2 sec` is `(pi)/(2) sqrt((1 + Npi^(2))` then what is the value of `N` ?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the speed of the particle at \( t = 2 \) seconds.
The particle starts from rest, so the initial speed \( u = 0 \). The speed increases at a constant rate of \( \frac{\pi}{2} \, \text{m/s}^2 \).

Using the formula for final velocity:
\[ v = u + at \]
Substituting the values:
\[ v = 0 + \left(\frac{\pi}{2}\right) \times 2 = \pi \, \text{m/s} \]

### Step 2: Calculate the centripetal acceleration.
Centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = \frac{v^2}{r} \]
Where \( r = 1 \, \text{m} \) (the radius of the circle).

Substituting the values:
\[ a_c = \frac{(\pi)^2}{1} = \pi^2 \, \text{m/s}^2 \]

### Step 3: Identify the tangential acceleration.
The tangential acceleration \( a_t \) is given as \( \frac{\pi}{2} \, \text{m/s}^2 \).

### Step 4: Calculate the net acceleration.
The net acceleration \( a \) can be found using the Pythagorean theorem:
\[ a = \sqrt{a_t^2 + a_c^2} \]
Substituting the values:
\[ a = \sqrt{\left(\frac{\pi}{2}\right)^2 + (\pi^2)^2} \]
Calculating each term:
\[ a = \sqrt{\frac{\pi^2}{4} + \pi^4} \]
Factoring out \( \pi^2 \):
\[ a = \sqrt{\pi^2 \left(\frac{1}{4} + \pi^2\right)} = \pi \sqrt{\frac{1}{4} + \pi^2} \]

### Step 5: Set the expression for net acceleration equal to the given expression.
We are given that the net acceleration at \( t = 2 \) seconds is:
\[ \frac{\pi}{2} \sqrt{1 + N\pi^2} \]
Setting the two expressions for net acceleration equal to each other:
\[ \pi \sqrt{\frac{1}{4} + \pi^2} = \frac{\pi}{2} \sqrt{1 + N\pi^2} \]

### Step 6: Simplify the equation.
Dividing both sides by \( \pi \):
\[ \sqrt{\frac{1}{4} + \pi^2} = \frac{1}{2} \sqrt{1 + N\pi^2} \]
Squaring both sides:
\[ \frac{1}{4} + \pi^2 = \frac{1}{4} (1 + N\pi^2) \]
Multiplying through by 4:
\[ 1 + 4\pi^2 = 1 + N\pi^2 \]
Subtracting 1 from both sides:
\[ 4\pi^2 = N\pi^2 \]
Dividing both sides by \( \pi^2 \) (assuming \( \pi \neq 0 \)):
\[ N = 4 \]

### Final Answer:
The value of \( N \) is \( 4 \).

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