A particle moves along the positive branch of the curve `y = (x^(2))/(2)` where `x = (t^(2))/(2),x` and y are measured in metres and t in second. At `t = 2s`, the velocity of the particle is

A particle moves along 1 ositive branch of the curve y=(x)/(2) where x=(t^(3))/(3) , x and y are measured in metres and t in seconds, then-

A particle moves along the positive branch of the curve y =2x2where x =2t2,where x and y are measured in metre and t in second. At t = 2 sec, the velocityof the particle is -

A particle moves along positive branch of the curve, y = x/2, where x = t^3/3 , x and y are measured in meters and t in seconds, then

A particle moves along the positive branch of the curve with x governed by where x and y are measured in meters and t in seconds. At t = 2s, the velocity of the particle is

A particle moves on the curve y = x^(2)/4 where x=t/2. x and y are meausured in metre and t in second. At t=4s the velocity of particel is

A particle moves along the curve y = x^2 /2. Here x varies with time as x = t^2 /2. Where x and y are measured in metres and t in seconds. At t = 2 s, the velocity of the particle (in ms^-1 ) is

A particle moves along the curve y = x^2 /2. Here x varies with time as x = t^2 /2. Where x and y are measured in metres and t in seconds. At t = 2 s, the velocity of the particle (in ms^-1 ) is

Completion type Question : The position coordinate of a moving particle is given by x = 6 + 18t + 9t^(2) (x is in metres and t in seconds) At t = 2s, the velocity of the particle will be …………

A particle moves along the curve y^(2) = 2x where x = (t^(2))/(2) & y gt 0 . What is the acceleration of the particle at t = 2s ?

A particle moves along the X-axis as x=u(t-2 s)+a(t-2 s)^2 .