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PhysicsKinematics2-D & 3DEasy2 minPYQ_2020

PhysicsEasysingle choice

The $x$ and $y$ coordinates of a particles at any time $t$ are given by $x = 7 t + 4 t^{2}$ and $y = 5 t$ . where $x$ and $y$ are in metre and $t$ in second. The acceleration of particle at $t = 5 s$ is

Options:

Answer:

Solution:

$a = \sqrt{a_{x}^{2} + a_{y}^{2}}$

$= {[{(\frac{d^{2} x}{{dt}^{2}})}^{2} + {(\frac{d^{2} y}{{dt}^{2}})}^{2}]}^{\frac{1}{2}}$

Hence, $a = \frac{d^{2} x}{d t^{2}} = 8 m s^{- 2}$

Stream:NTA_ABHYASSubject:PhysicsTopic:KinematicsSubtopic:2-D & 3D

⏱ 2mℹ️ Source: PYQ_2020

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