The velocity varies with time as `vecv = ahati + bthatj` , where a and b are positive constants. The magnitude of instantaneous velocity and acceleration would be

Options

A. `a^2+b^2t^2,b`
B. `sqrt(a^2+b^2t^2),a`
C. `sqrt(a^2+b^2t^2),b`
D. `a^2+b^2t^2,a`

Answer

To find the magnitude of instantaneous velocity and acceleration given the velocity vector \(\vec{v} = a \hat{i} + b t \hat{j}\), where \(a\) and \(b\) are positive constants, we can follow these steps:

### Step 1: Identify the components of the velocity vector
The velocity vector is given as:
\[ \vec{v} = a \hat{i} + b t \hat{j} \]
Here, the components are:
- \(v_x = a\) (the coefficient of \(\hat{i}\))
- \(v_y = b t\) (the coefficient of \(\hat{j}\))

### Step 2: Calculate the magnitude of the instantaneous velocity
The magnitude of a vector \(\vec{v} = v_x \hat{i} + v_y \hat{j}\) is given by:
\[ |\vec{v}| = \sqrt{v_x^2 + v_y^2} \]
Substituting the components of the velocity:
\[ |\vec{v}| = \sqrt{a^2 + (b t)^2} \]
Thus, the magnitude of the instantaneous velocity is:
\[ |\vec{v}| = \sqrt{a^2 + b^2 t^2} \]

### Step 3: Find the instantaneous acceleration
The acceleration \(\vec{a}\) is the derivative of the velocity vector with respect to time:
\[ \vec{a} = \frac{d\vec{v}}{dt} \]
Differentiating each component:
- The derivative of \(a\) (a constant) is \(0\).
- The derivative of \(b t\) with respect to \(t\) is \(b\).

Thus, the acceleration vector is:
\[ \vec{a} = 0 \hat{i} + b \hat{j} = b \hat{j} \]

### Step 4: Calculate the magnitude of the instantaneous acceleration
The magnitude of the acceleration vector is:
\[ |\vec{a}| = \sqrt{(0)^2 + (b)^2} = \sqrt{b^2} = b \]

### Final Results
- The magnitude of instantaneous velocity is:
\[ |\vec{v}| = \sqrt{a^2 + b^2 t^2} \]
- The magnitude of instantaneous acceleration is:
\[ |\vec{a}| = b \]