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

Explanation

To find the magnitude of the acceleration of the particle at time \( t = 1 \), we will follow these steps:

### Step 1: Write the position vector
The position vector of the particle is given by:
\[ \vec{r}(t) = 15t^2 \hat{i} + (4 - 20t^2) \hat{j} \]

### Step 2: Differentiate the position vector to find the velocity
The velocity \(\vec{v}(t)\) is the first derivative of the position vector with respect to time:
\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(15t^2 \hat{i} + (4 - 20t^2) \hat{j}) \]
Differentiating each component:
- For the \( \hat{i} \) component:
\[ \frac{d}{dt}(15t^2) = 30t \]
- For the \( \hat{j} \) component:
\[ \frac{d}{dt}(4 - 20t^2) = -40t \]
Thus, the velocity vector becomes:
\[ \vec{v}(t) = 30t \hat{i} - 40t \hat{j} \]

### Step 3: Differentiate the velocity vector to find the acceleration
The acceleration \(\vec{a}(t)\) is the derivative of the velocity vector with respect to time:
\[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(30t \hat{i} - 40t \hat{j}) \]
Differentiating each component:
- For the \( \hat{i} \) component:
\[ \frac{d}{dt}(30t) = 30 \]
- For the \( \hat{j} \) component:
\[ \frac{d}{dt}(-40t) = -40 \]
Thus, the acceleration vector becomes:
\[ \vec{a}(t) = 30 \hat{i} - 40 \hat{j} \]

### Step 4: Find the magnitude of the acceleration vector
The magnitude of the acceleration vector is given by:
\[ |\vec{a}| = \sqrt{(30)^2 + (-40)^2} \]
Calculating the squares:
\[ |\vec{a}| = \sqrt{900 + 1600} = \sqrt{2500} \]
Thus, the magnitude of the acceleration is:
\[ |\vec{a}| = 50 \]

### Final Answer
The magnitude of the acceleration at \( t = 1 \) is \( 50 \).

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