At time t = 0 a particle starts travelling from a height 7z cm in a plane keeping z coordinate constant. At any instant of time its position along the x and y directions are defined as 3t and 5t³ respectively. At t = 1s acceleration of the particle will be (a) -30y (b) 30y (c) 3x + 15y (d) 3x + 15y + 7z

Answer: (b). x = 3t ⇒ ax = 0; z = 7 constant ⇒ az = 0 y = 5t³ ⇒ ay = 30t = 30 at t = 1 s So the answer is (b) 30 y.

At time t = 0 a particle starts travelling from a height 7z cm in a… | JEE Main 2022 (28 Jul, Shift 2)

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Motion in a Plane (2-D Kinematics)

At time $t = 0$ a particle starts travelling from a height $7\hat{z}\text{ cm}$ in a plane keeping z coordinate constant. At any instant of time its position along the $\hat{x}$ and $\hat{y}$ directions are defined as $3t$ and $5t^{3}$ respectively. At t = 1s acceleration of the particle will be

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Answer: (b)

$x = 3t \Rightarrow a_x = 0$ ; $z = 7$ constant $\Rightarrow a_z = 0$
$y = 5t^3 \Rightarrow a_y = 30t = 30$ at $t = 1$ s
So the answer is (b) $30\hat y$ .

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