KinematicsHard
Question
If velocity of the particle is given by v = 
, where x denotes position of the particle and initially particle was at x = 4m, then which of the following are correct.
, where x denotes position of the particle and initially particle was at x = 4m, then which of the following are correct.Options
A.At t = 2 s, the position of the particle is at x = 9m
B.Particle acceleration at t = 2 s. is 1 m/s2
C.Particle acceleration is 1 / 2 m/s2 through out the motion
D.Particle will never go in negative direction from it’s staring position
Solution
v = 
; 
⇒ [2
]4x = t
⇒ x =
at t = 2 ⇒ x = 9m
a = v
= 
× 
= 
m/s2
at = 4 ⇒ v = 2m/s & it increases as x increases so it never becomes negative.
; ⇒ [2]4x = t⇒ x =
at t = 2 ⇒ x = 9ma = v
= × = m/s2at = 4 ⇒ v = 2m/s & it increases as x increases so it never becomes negative.
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