At time t = 0 a particle starts travelling from a height 7 z ^ cm in a plane keeping z coordinate constant. At any instant of time, it's position along the x and y directions are defined as 3 t a

Question

At time t = 0 a particle starts travelling from a height 7 z ^ cm in a plane keeping z coordinate constant. At any instant of time, it's position along the x and y directions are defined as 3 t and 5 t 3 respectively. At t = 1 s acceleration of the particle will be

TestHub · Accepted Answer

The position vector of the particle can be written as,
$\vec{r} = 3 t \hat{i} + 5 t^{3} \hat{j} + 7 \hat{k}$
The velocity of the particle will be,
$\vec{v} = \frac{d \vec{r}}{d t} = 3 \hat{i} + 15 t^{2} \hat{j}$
Now the acceleration of the particle will be,
$\vec{a} = \frac{d \vec{v}}{d t} = \frac{d^{2} \vec{r}}{d t^{2}} = 30 t \hat{j}$
At $t = 1 s \Rightarrow \frac{d^{2} r}{d t^{2}} = 30 \hat{j}$

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