A particle leaves the origin with an initial velocity v → = 3.00 i ^ ms - 1 and a constant acceleration a → = - 1.00 i ^ - 0.5 j ^ ms - 2 . When the particle reaches it maximum x -coordinate, what is the magnitude of its velocity (in m/s) in y -direction?

Velocity along x direction is v x = v 0 2 + at v x = 3 + - 1 t so when particle reach maximum x coordinate v x = 0 ⇒ t = 3 sec similarly v y = v 0 y + a y t v y = 0 + - 1 2 t v y at t = 3 = - 3 2 = - 1 . 5 m s - 1

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PhysicsKinematics2-D & 3DEasy2 minPYQ_2020

PhysicsEasynumerical

A particle leaves the origin with an initial velocity $\vec{v} = (3.00 \hat{i}) {ms}^{- 1}$ and a constant acceleration $\vec{a} = (- 1.00 \hat{i} - 0.5 \hat{j}) {ms}^{- 2}$ . When the particle reaches it maximum $x$ -coordinate, what is the magnitude of its velocity (in m/s) in $y$ -direction?

Answer:

1.50

Solution:

Velocity along $x$ direction is

$v_{x} = v_{02} + at$

$v_{x} = 3 + (- 1) t$

so when particle reach maximum $x$ coordinate $v_{x} = 0 \Rightarrow t = 3 sec$

similarly $v_{y} = v_{0 y} + a_{y} t$

$v_{y} = 0 + (- \frac{1}{2}) t$

${(v_{y})}_{at t = 3} = - \frac{3}{2} = - 1.5 m s^{- 1}$

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

⏱ 2mℹ️ Source: PYQ_2020

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