JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 2)
At time $$t=0$$ a particle starts travelling from a height $$7 \hat{z} \mathrm{~cm}$$ in a plane keeping z coordinate constant. At any instant of time it's position along the $$\hat{x}$$ and $$\hat{y}$$ directions are defined as $$3 \mathrm{t}$$ and $$5 \mathrm{t}^{3}$$ respectively. At t = 1s acceleration of the particle will be
$$-30 \hat{y}$$
$$30 \hat{y}$$
$$3 \hat{x}+15 \hat{y}$$
$$3 \hat{x}+15 \hat{y}+7 \hat{z}$$
Explanation
$$x = 3t \Rightarrow {a_x} = 0$$
$$y = 5{t^3} \Rightarrow {a_y} = 30t$$
$$\overrightarrow a (t = 1) = 30\widehat y$$
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