Consider a system of two vector `overset(rarr)A` and `overset(rarr)B` changing with respect to time `(t ge 0)`. `overset(rarr)a=8 t hat i - t^(2) hatj, overset(rarr)b=t^(2)hat i+2 hat j` ,

Options

A. The vector will become parallel to each other at t = 4s
B. The vectors will never become perpendicular to each other
C. The vectors will become perpendicular to each other at t = 1/4 s
D. The vector will become parallel to each other at t = 2 s

Explanation

To solve the problem, we need to analyze the vectors \(\overset{\rarr}{a}\) and \(\overset{\rarr}{b}\) given as:

\[ \overset{\rarr}{a} = 8t \hat{i} - t^2 \hat{j} \]

\[ \overset{\rarr}{b} = t^2 \hat{i} + 2 \hat{j} \]

We will check if these vectors become parallel or perpendicular at any given time \(t\).

### Step 1: Check for Parallelism

Vectors are parallel if one is a scalar multiple of the other. This means we need to check if there exists a constant \(k\) such that:

\[ \overset{\rarr}{a} = k \overset{\rarr}{b} \]

### Step 2: Calculate the Vectors at Specific Times

#### Check at \(t = 4\):

1. Calculate \(\overset{\rarr}{a}\):
\[ \overset{\rarr}{a} = 8(4) \hat{i} - (4^2) \hat{j} = 32 \hat{i} - 16 \hat{j} \]

2. Calculate \(\overset{\rarr}{b}\):
\[ \overset{\rarr}{b} = (4^2) \hat{i} + 2 \hat{j} = 16 \hat{i} + 2 \hat{j} \]

Now, we need to check if \(\overset{\rarr}{a} = k \overset{\rarr}{b}\) for some \(k\).

From the \(i\) components:
\[ 32 = 16k \implies k = 2 \]

From the \(j\) components:
\[ -16 = 2k \implies k = -8 \]

Since \(k\) is not the same for both components, the vectors are not parallel at \(t = 4\).

#### Check at \(t = 2\):

1. Calculate \(\overset{\rarr}{a}\):
\[ \overset{\rarr}{a} = 8(2) \hat{i} - (2^2) \hat{j} = 16 \hat{i} - 4 \hat{j} \]

2. Calculate \(\overset{\rarr}{b}\):
\[ \overset{\rarr}{b} = (2^2) \hat{i} + 2 \hat{j} = 4 \hat{i} + 2 \hat{j} \]

Again, check if \(\overset{\rarr}{a} = k \overset{\rarr}{b}\).

From the \(i\) components:
\[ 16 = 4k \implies k = 4 \]

From the \(j\) components:
\[ -4 = 2k \implies k = -2 \]

Since \(k\) is not the same for both components, the vectors are not parallel at \(t = 2\).

### Step 3: Check for Perpendicularity

Vectors are perpendicular if their dot product is zero:

\[ \overset{\rarr}{a} \cdot \overset{\rarr}{b} = 0 \]

Calculating the dot product:

\[ \overset{\rarr}{a} \cdot \overset{\rarr}{b} = (8t)(t^2) + (-t^2)(2) \]
\[ = 8t^3 - 2t^2 \]

Set the dot product to zero:

\[ 8t^3 - 2t^2 = 0 \]

Factor out \(2t^2\):

\[ 2t^2(4t - 1) = 0 \]

This gives us:

1. \(t^2 = 0 \implies t = 0\)
2. \(4t - 1 = 0 \implies t = \frac{1}{4}\)

### Conclusion

The vectors \(\overset{\rarr}{a}\) and \(\overset{\rarr}{b}\) are perpendicular at \(t = 0\) and \(t = \frac{1}{4}\) seconds. They are never parallel at \(t = 2\) or \(t = 4\).