A particle whose speed is `50ms^(-1)` moves along the line from `A(2,1)` to `B(9,25)`. Find its velocity vector in the from of `ahat(i)+bhat(j)`.

Options

A. `(7hat(i) + 24hat(j))m//s`
B. `2(7hat(i) + 24hat(j))m//s`
C. `4(7hat(i) + 24hat(j))m//s`
D. `5(7hat(i) + 24hat(j))m//s`

Answer

To find the velocity vector of a particle moving from point A(2,1) to point B(9,25) with a speed of 50 m/s, we can follow these steps:

### Step 1: Determine the position vectors of points A and B
The position vector of point A can be represented as:
\[ \vec{OA} = 2\hat{i} + 1\hat{j} \]
The position vector of point B can be represented as:
\[ \vec{OB} = 9\hat{i} + 25\hat{j} \]

### Step 2: Find the displacement vector from A to B
The displacement vector \(\vec{AB}\) can be calculated as:
\[ \vec{AB} = \vec{OB} - \vec{OA} = (9\hat{i} + 25\hat{j}) - (2\hat{i} + 1\hat{j}) = (9 - 2)\hat{i} + (25 - 1)\hat{j} = 7\hat{i} + 24\hat{j} \]

### Step 3: Calculate the magnitude of the displacement vector
The magnitude of the displacement vector \(\vec{AB}\) is given by:
\[ |\vec{AB}| = \sqrt{(7^2) + (24^2)} = \sqrt{49 + 576} = \sqrt{625} = 25 \]

### Step 4: Determine the unit vector in the direction of AB
The unit vector \(\hat{AB}\) in the direction of \(\vec{AB}\) is calculated as:
\[ \hat{AB} = \frac{\vec{AB}}{|\vec{AB}|} = \frac{7\hat{i} + 24\hat{j}}{25} = \frac{7}{25}\hat{i} + \frac{24}{25}\hat{j} \]

### Step 5: Calculate the velocity vector
The velocity vector \(\vec{v}\) can be expressed as:
\[ \vec{v} = \text{speed} \times \hat{AB} = 50 \times \left(\frac{7}{25}\hat{i} + \frac{24}{25}\hat{j}\right) \]
Calculating this gives:
\[ \vec{v} = 50 \times \frac{7}{25}\hat{i} + 50 \times \frac{24}{25}\hat{j} = 2 \times 7\hat{i} + 2 \times 24\hat{j} = 14\hat{i} + 48\hat{j} \]

### Final Result
Thus, the velocity vector in the form of \(a\hat{i} + b\hat{j}\) is:
\[ \vec{v} = 14\hat{i} + 48\hat{j} \]

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