Position of particle during motion is expressed as $\vec{r} = (5t^2\hat{i} + 3t \hat{j})$ m. The magnitude of acceleration of particle at $t = 0$, is

Solution:

Given the position vector of the particle as $\vec{r} = (5t^2\hat{i} + 3t \hat{j})$ m.

To find the velocity vector, we differentiate the position vector with respect to time:

$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(5t^2\hat{i} + 3t \hat{j})$

$\vec{v} = (10t\hat{i} + 3 \hat{j})$ m/s

To find the acceleration vector, we differentiate the velocity vector with respect to time:

$\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(10t\hat{i} + 3 \hat{j})$

$\vec{a} = (10\hat{i} + 0 \hat{j})$ m/s²

$\vec{a} = 10\hat{i}$ m/s²

The acceleration vector is constant and does not depend on time $t$. Therefore, the acceleration at $t=0$ is the same as at any other time.

The magnitude of the acceleration is:

$|\vec{a}| = \sqrt{(10)^2 + (0)^2} = \sqrt{100} = 10$ m/s²

Thus, the magnitude of acceleration of the particle at $t=0$ is 10 m/s².