A point moves according to the `law" "x= at, y=at (1 -alphat)` where a and `alpha` are positive constants and t is time . If the moment at which angle between velocity vecotrs and acceleration vectors is `(pi)/(4)` is given by `(A)/(alpha)`. Find the value of A .

Explanation

To solve the problem, we need to analyze the motion of the point described by the equations \( x = at \) and \( y = at(1 - \alpha t) \). We will find the time \( t \) when the angle between the velocity vector and acceleration vector is \( \frac{\pi}{4} \).

### Step 1: Find the velocity vector
The velocity vector \( \vec{v} \) is given by the derivatives of \( x \) and \( y \) with respect to time \( t \).

\[ \vec{v} = \left( \frac{dx}{dt}, \frac{dy}{dt} \right) \]

Calculating the derivatives:
- \( \frac{dx}{dt} = a \)
- \( \frac{dy}{dt} = a(1 - 2\alpha t) \)

Thus, the velocity vector is:
\[ \vec{v} = a \hat{i} + a(1 - 2\alpha t) \hat{j} \]

### Step 2: Find the acceleration vector
The acceleration vector \( \vec{a} \) is given by the second derivatives of \( x \) and \( y \) with respect to time \( t \).

\[ \vec{a} = \left( \frac{d^2x}{dt^2}, \frac{d^2y}{dt^2} \right) \]

Calculating the second derivatives:
- \( \frac{d^2x}{dt^2} = 0 \)
- \( \frac{d^2y}{dt^2} = -2a\alpha \)

Thus, the acceleration vector is:
\[ \vec{a} = 0 \hat{i} - 2a\alpha \hat{j} \]

### Step 3: Find the angle between the velocity and acceleration vectors
The angle \( \theta \) between two vectors can be found using the dot product formula:
\[ \vec{v} \cdot \vec{a} = |\vec{v}| |\vec{a}| \cos(\theta) \]

Given that \( \theta = \frac{\pi}{4} \), we have:
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Step 4: Calculate the magnitudes of \( \vec{v} \) and \( \vec{a} \)
The magnitude of the velocity vector \( |\vec{v}| \) is:
\[ |\vec{v}| = \sqrt{a^2 + (a(1 - 2\alpha t))^2} = a \sqrt{1 + (1 - 2\alpha t)^2} \]

The magnitude of the acceleration vector \( |\vec{a}| \) is:
\[ |\vec{a}| = \sqrt{0^2 + (-2a\alpha)^2} = 2a\alpha \]

### Step 5: Set up the equation using the dot product
The dot product \( \vec{v} \cdot \vec{a} \) is:
\[ \vec{v} \cdot \vec{a} = a(0) + a(1 - 2\alpha t)(-2a\alpha) = -2a^2\alpha(1 - 2\alpha t) \]

Using the angle condition:
\[ -2a^2\alpha(1 - 2\alpha t) = |\vec{v}| |\vec{a}| \cos\left(\frac{\pi}{4}\right) \]
\[ -2a^2\alpha(1 - 2\alpha t) = a \sqrt{1 + (1 - 2\alpha t)^2} \cdot 2a\alpha \cdot \frac{1}{\sqrt{2}} \]

### Step 6: Simplify and solve for \( t \)
After simplifying, we will find \( t \) when the angle is \( \frac{\pi}{4} \):
\[ -2(1 - 2\alpha t) = \sqrt{1 + (1 - 2\alpha t)^2} \]

Squaring both sides and solving will give us the value of \( t \).

### Step 7: Find \( A \)
The problem states that the moment when the angle is \( \frac{\pi}{4} \) is given by \( \frac{A}{\alpha} \). After solving for \( t \), we can express it in the form \( \frac{A}{\alpha} \) and find \( A \).

### Final Result
After solving, we find that \( A = 1 \).

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