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Step-by-Step Solution
Step 1: Write down the position vector
The position vector of the particle is given by:
$ \overrightarrow{r}(t) = 15t^2 \,\hat{i} + \bigl(4 - 20t^2\bigr)\,\hat{j} $
Step 2: Differentiate to find the velocity
Velocity is the first derivative of the position vector with respect to time:
$ \overrightarrow{v}(t) = \frac{d\overrightarrow{r}(t)}{dt} = \frac{d}{dt}\bigl(15t^2 \,\hat{i} + (4 - 20t^2)\,\hat{j}\bigr). $
So,
$ \overrightarrow{v}(t) = 30t \,\hat{i} - 40t \,\hat{j}. $
Step 3: Differentiate to find the acceleration
Acceleration is the second derivative of the position vector (or the first derivative of velocity) with respect to time:
$ \overrightarrow{a}(t) = \frac{d\overrightarrow{v}(t)}{dt} = \frac{d}{dt}(30t \,\hat{i} - 40t \,\hat{j}). $
Therefore,
$ \overrightarrow{a}(t) = 30 \,\hat{i} - 40 \,\hat{j}. $
Step 4: Evaluate acceleration at t = 1
Substitute $ t = 1 $ into $ \overrightarrow{a}(t) $:
$ \overrightarrow{a}(1) = 30 \,\hat{i} - 40 \,\hat{j}. $
Step 5: Calculate the magnitude
The magnitude of the acceleration vector is given by:
$ \bigl|\overrightarrow{a}(1)\bigr| = \sqrt{(30)^2 + (-40)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50. $
Final Answer
The magnitude of the acceleration at $ t = 1 $ is 50.
