KinematicsHard
Question
A particle leaves the origin an initial velocity 
= 3
m/s and a constant acceleration 
= (-1.0î - 0.5ĵ) m/s2 Its velocity 
and position vector 
when it reaches its maximum x-co-ordinate are :-
= 3 m/s and a constant acceleration = (-1.0î - 0.5ĵ) m/s2 Its velocity and position vector when it reaches its maximum x-co-ordinate are :- Options
A. = - 2ĵ
= - 2ĵB. = - 1.5ĵ m/s
= - 1.5ĵ m/sC.
= - (4.5î - 2.25ĵ) m
= - (4.5î - 2.25ĵ) mD. = - (3î - 2ĵ) m
= - (3î - 2ĵ) mSolution
(t) = (3 - 1 × t) î + (0 - 0.5) ĵ ..............(i)For maximum positive x coordinate when vx becomes zero
∴ 3 - t = 0 ⇒ t = 3 sec
then
(3) = 4.5 î - 2.25 ĵCreate a free account to view solution
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