At time t = 0,a particle starts travelling fr | vedclass

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(B) The position vector of the particle at any time $t$ is given by $\vec{r}(t) = x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k}$.Given $x = 3t$,$y = 5t^3

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$30 \, \hat{j} \, 	ext{cm/s}^2$

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(B) The position vector of the particle at any time $t$ is given by $\vec{r}(t) = x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k}$.Given $x = 3t$,$y = 5t^3$,and $z = 7$ (constant).So,$\vec{r}(t) = 3t \hat{i} + 5t^3 \hat{j} + 7 \hat{k}$.The velocity vector $\vec{v}$ is the first derivative of position with respect to time: $\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3t) \hat{i} + \frac{d}{dt}(5t^3) \hat{j} + \frac{d}{dt}(7) \hat{k} = 3 \hat{i} + 15t^2 \hat{j}$.The acceleration vector $\vec{a}$ is the derivative of velocity with respect to time: $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(3) \hat{i} + \frac{d}{dt}(15t^2) \hat{j} = 0 \hat{i} + 30t \hat{j}$.At $t = 1 \, 	ext{s}$,the acceleration is $\vec{a} = 30(1) \hat{j} = 30 \hat{j} \, 	ext{cm/s}^2$.

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