A particle moves in the XY-plane according to the law x = k t , y = k t ( 1 − α t ) , where k and α are positive constants and t is time find time tnod after which the angle between velocity and acceleration vector is 45

1. Position–time relation

The particle’s coordinates are given by
$x = k t, \qquad y = k t \bigl(1 - \alpha t\bigr)$ where $k$ and $\alpha$ are positive constants.

2. Velocity vector $\vec v$

Velocity is the time derivative of position.

$$\vec v = \frac{d\vec r}{dt} = \Bigl(\frac{dx}{dt},\frac{dy}{dt}\Bigr) = \bigl(k,\; k(1 - 2\alpha t)\bigr)$$

Interpretation:

• $k$ in $x$-direction: constant horizontal speed.
• $k(1-2\alpha t)$ in $y$-direction: vertical speed that decreases linearly with time.

3. Acceleration vector $\vec a$

Acceleration is the derivative of velocity.

$$\vec a = \frac{d\vec v}{dt} = \bigl(0,\; -2k\alpha\bigr)$$

So acceleration is a constant vector pointing straight down (negative $y$). No horizontal acceleration.

4. Angle between two vectors

For any two vectors $\vec p$ and $\vec q$ the cosine of the angle $\theta$ between them is

$$\cos\theta = \frac{\vec p\cdot\vec q}{\lVert\vec p\rVert\,\lVert\vec q\rVert}$$

We need $\theta = 45^{\circ}$, so $\cos\theta = \frac{1}{\sqrt 2}$.

5. Substitute $\vec v$ and $\vec a$

Dot product:

$$\vec v\cdot\vec a = k \cdot 0 + k(1 - 2\alpha t)(-2k\alpha) = -2k^2\alpha\bigl(1 - 2\alpha t\bigr)$$

Magnitudes:

$$\lVert\vec v\rVert = k\sqrt{1 + \bigl(1 - 2\alpha t\bigr)^2},\qquad \lVert\vec a\rVert = 2k\alpha$$

6. Apply the angle condition

$$\frac{-2k^2\alpha\bigl(1 - 2\alpha t\bigr)}{\bigl[k\sqrt{1 + (1 - 2\alpha t)^2}\bigr]\,(2k\alpha)} = \frac{1}{\sqrt 2}$$

After cancelling common factors $k, 2k\alpha$ we get

$$-\frac{1 - 2\alpha t}{\sqrt{1 + (1 - 2\alpha t)^2}} = \frac{1}{\sqrt 2}$$

Let $s = 1 - 2\alpha t$.

Equation becomes

$$-\frac{s}{\sqrt{1 + s^2}} = \frac{1}{\sqrt 2}$$

Square both sides:

$$\frac{s^2}{1 + s^2} = \frac12$$

Solve for $s$:

$$2s^2 = 1 + s^2 \quad\Rightarrow\quad s^2 = 1$$

Because of the leading minus sign, we need $s = -1$ (to keep left side positive):

$$1 - 2\alpha t = -1 \;\Rightarrow\; 2\alpha t = 2 \;\Rightarrow\; t = \frac{1}{\alpha}$$

That is the required time.

Think of the particle like an ant walking on a sheet of paper

1. Where is the ant?
   At any time $t$, its place is decided by two rules:
   $x = k t$ (how far right) and $y = k t (1 - \alpha t)$ (how far up).

2. How fast is it moving?
   Velocity tells us the direction and speed of the ant at that moment.

3. How is its speed changing?
   Acceleration says whether the ant is speeding up or slowing down and in which direction.

4. Angle between velocity and acceleration = $45^{\circ}$
   We want to know when the direction of velocity makes a $45^{\circ}$ angle with the direction of acceleration.
   It is like asking: At what time do the two arrows (velocity-arrow and acceleration-arrow) form half of a right angle?

5. Answer turns out to be
   $t = \frac{1}{\alpha}$

That is the special time when the ant's direction and the way its speed is changing are exactly $45^{\circ}$ apart.

Step-by-Step Solution

1. Write position vector
   $\vec r = \bigl(k t\bigr)\,\hat i + \bigl(k t (1 - \alpha t)\bigr)\,\hat j$

2. Velocity
   $\vec v = \frac{d\vec r}{dt} = k\,\hat i + k(1 - 2\alpha t)\,\hat j$

3. Acceleration
   $\vec a = \frac{d\vec v}{dt} = 0\,\hat i - 2k\alpha\,\hat j$

4. Dot product
   $\vec v\cdot\vec a = -2k^2\alpha\bigl(1 - 2\alpha t\bigr)$

5. Magnitudes
   $\lVert\vec v\rVert = k\sqrt{1 + \bigl(1 - 2\alpha t\bigr)^2}, \qquad \lVert\vec a\rVert = 2k\alpha$

6. Angle condition ($\theta = 45^{\circ}$)
   $\cos45^{\circ} = \frac{\vec v\cdot\vec a}{\lVert\vec v\rVert \lVert\vec a\rVert} = \frac{1}{\sqrt2}$

   Substituting values and simplifying:

   $-\frac{1 - 2\alpha t}{\sqrt{1 + (1 - 2\alpha t)^2}} = \frac{1}{\sqrt2}$

7. Solve for $t$

   Let $s = 1 - 2\alpha t$.

   $-\frac{s}{\sqrt{1 + s^2}} = \frac{1}{\sqrt2} \;\Longrightarrow\; \frac{s^2}{1 + s^2} = \frac12$

   $2s^2 = 1 + s^2 \;\Rightarrow\; s^2 = 1 \;\Rightarrow\; s = -1$

   $1 - 2\alpha t = -1 \;\Rightarrow\; 2\alpha t = 2 \;\Rightarrow\; t = \frac{1}{\alpha}$

8. Final answer
   $\boxed{t_{\text{nod}} = \dfrac{1}{\alpha}}$