Kinematics

JEE Main level, single correct, medium

186 of 308

Question 186 of 308

At time t=0t=0 a particle starts travelling from a height 7z^cm7\hat{z}cm in a plane keeping zz coordinate constant. At any instant of time, it's position along the xx and yy directions are defined as 3t3t and 5t35{t}^{3} respectively. At t=1st=1s acceleration of the particle will be

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Answer: (B) 30y30y

The position vector of the particle can be written as,
r⃗=3ti^+5t3j^+7k^\vec{r}=3t\hat{i}+5{t}^{3}\hat{j}+7\hat{k}
The velocity of the particle will be,
v⃗=dr⃗dt=3i^+15t2j^\vec{v}=\dfrac{d\vec{r}}{dt}=3\hat{i}+15{t}^{2}\hat{j}
Now the acceleration of the particle will be,
a⃗=dv⃗dt=d2r⃗dt2=30tj^\vec{a}=\dfrac{d\vec{v}}{dt}=\dfrac{{d}^{2}\vec{r}}{d{t}^{2}}=30t\hat{j}
At t=1s⇒d2rdt2=30j^t=1s\Rightarrow \dfrac{{d}^{2}r}{d{t}^{2}}=30\hat{j}

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