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

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-30 $\hat{y}$

30 $\hat{y}$

3 $\hat{x}$ + 15 $\hat{y}$

$3 \hat{x} + 15 \hat{y} + 7 \overset{\land}{z}$

answer is B.

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Detailed Solution

$\overset{⇀}{r} = 3 t \hat{i} + 5 t^{3} \hat{j} + 7 \hat{k}$

$\frac{d^{2} \overset{⇀}{r}}{{dt}^{2}} = 30 t \hat{j} At t = 1$

$\Rightarrow \frac{d^{2} \overset{⇀}{r}}{{dt}^{2}} = 30 \hat{j}$

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