The position vector of a particle changes with time according to the direction `vec(r)(t)=15 t^(2)hati+(4-20 t^(2))hatj`. What is the magnitude of the acceleration at t=1?

Options

A. 40
B. 25
C. 100
D. 50

Answer

To solve the problem, we need to find the magnitude of the acceleration of a particle whose position vector is given by:

\[ \vec{r}(t) = 15t^2 \hat{i} + (4 - 20t^2) \hat{j} \]

### Step 1: Differentiate the position vector to find the velocity vector.

The velocity vector \(\vec{v}(t)\) is given by the derivative of the position vector with respect to time:

\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(15t^2 \hat{i} + (4 - 20t^2) \hat{j}) \]

Calculating the derivative:

\[ \vec{v}(t) = \frac{d}{dt}(15t^2) \hat{i} + \frac{d}{dt}(4 - 20t^2) \hat{j} \]

\[ \vec{v}(t) = 30t \hat{i} - 40t \hat{j} \]

### Step 2: Differentiate the velocity vector to find the acceleration vector.

The acceleration vector \(\vec{a}(t)\) is given by the derivative of the velocity vector with respect to time:

\[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(30t \hat{i} - 40t \hat{j}) \]

Calculating the derivative:

\[ \vec{a}(t) = \frac{d}{dt}(30t) \hat{i} + \frac{d}{dt}(-40t) \hat{j} \]

\[ \vec{a}(t) = 30 \hat{i} - 40 \hat{j} \]

### Step 3: Find the magnitude of the acceleration vector.

The magnitude of the acceleration vector \(\vec{a}\) is given by:

\[ |\vec{a}| = \sqrt{(30)^2 + (-40)^2} \]

Calculating the squares:

\[ |\vec{a}| = \sqrt{900 + 1600} = \sqrt{2500} \]

Calculating the square root:

\[ |\vec{a}| = 50 \]

### Conclusion

The magnitude of the acceleration at \(t = 1\) second is:

\[ \boxed{50} \]