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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Motion in One Dimension

NTA Abhyas 2020 · Physics · Motion in One Dimension

A particle leaves the origin with an initial velocity → v = (3.00 ^...

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

Correct answer

1.5

Step-by-step solution

Velocity along xx direction is

vx=v02+at{v}_{x} = {v}_{0 2} + \text{at}

vx=3+(−1)t{v}_{x} = 3 + \left(- 1\right) t

so when particle reach maximum xx coordinate vx=0⇒t=3sec{v}_{x} = 0 ⇒ t = 3 sec

similarly vy=v0y+ayt{v}_{y} = {v}_{0 y} + {a}_{y} t

vy=0+(−12)t{v}_{y} = 0 + \left(- \dfrac{1}{2}\right) t

(vy)att=3=−32=−1.5ms−1{\left({v}_{y}\right)}_{at t = 3} = - \dfrac{3}{2} = - 1 . 5m{s}^{- 1}
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