The position vector of a aprticle is given as `vecr=(t^2-4t+6)hati+(t^2)hatj`. The time after which the velocity vector and acceleration vector becomes perpendicular to each other is equal to

Options

A. `1 sec`
B. `2 sec`
C. `1.5 sec`
D. not possible

Explanation

To solve the problem, we need to find the time at which the velocity vector and acceleration vector of the particle become perpendicular to each other. This can be determined when their dot product equals zero.

### Step 1: Write down the position vector
The position vector \( \vec{r} \) is given as:
\[ \vec{r} = (t^2 - 4t + 6) \hat{i} + (t^2) \hat{j} \]

### Step 2: Find the velocity vector
The velocity vector \( \vec{v} \) is the derivative of the position vector with respect to time \( t \):
\[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}[(t^2 - 4t + 6) \hat{i} + (t^2) \hat{j}] \]
Calculating the derivatives:
- For the \( \hat{i} \) component:
\[ \frac{d}{dt}(t^2 - 4t + 6) = 2t - 4 \]
- For the \( \hat{j} \) component:
\[ \frac{d}{dt}(t^2) = 2t \]
Thus, the velocity vector is:
\[ \vec{v} = (2t - 4) \hat{i} + (2t) \hat{j} \]

### Step 3: Find the acceleration vector
The acceleration vector \( \vec{a} \) is the derivative of the velocity vector with respect to time \( t \):
\[ \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}[(2t - 4) \hat{i} + (2t) \hat{j}] \]
Calculating the derivatives:
- For the \( \hat{i} \) component:
\[ \frac{d}{dt}(2t - 4) = 2 \]
- For the \( \hat{j} \) component:
\[ \frac{d}{dt}(2t) = 2 \]
Thus, the acceleration vector is:
\[ \vec{a} = 2 \hat{i} + 2 \hat{j} \]

### Step 4: Set up the dot product condition
To find when the vectors are perpendicular, we set their dot product to zero:
\[ \vec{v} \cdot \vec{a} = 0 \]
Calculating the dot product:
\[ (2t - 4) \cdot 2 + (2t) \cdot 2 = 0 \]
This simplifies to:
\[ 4t - 8 + 4t = 0 \]
Combining like terms:
\[ 8t - 8 = 0 \]

### Step 5: Solve for \( t \)
Now, we can solve for \( t \):
\[ 8t = 8 \implies t = 1 \text{ second} \]

### Conclusion
The time after which the velocity vector and acceleration vector become perpendicular is:
\[ \boxed{1 \text{ second}} \]