Given `a=-cost,` at `t=0,u=0,x=1`
Position at `t=pi`
Find distance from 0 to `2pi`.

A particle is moving with an acceleration given by, a=-cos t ; where time t is in seconds and a is in m/s ^(2) . It is given that at t=0, u=0 and x=1 .Find distance travelled by particle from pi to 2 pi .

Given a=4-2t , initial velocity at t=0,u=5, find distance travelled till 12 sec.

A particle executes simple harmonic motion about x=0 along x-axis. The position of the particle at an instant is given by x= ( 5 cm) sin pi t . Find the average velocity and average acceleration for a time interval 0-0.5 s . Hint : Average velocity =( x_(2)- x_(1))/(t_(2)-t_(1)) x_(1) = 5 sin 0 x_(2) = 5 sin ( pi xx 0.5) t_(2) - t_(1) = 0.5 Average acceleration = ( v_(2) - v_(1))/( t_(2) - t_(1)) v=(dx)/(dt) = 5 pi cos pit implies a _(av) = ( 5 pi cos ( pi xx 0.5) -5 pi cos theta)/(0.5)

A particle executes simple harmonic motion about x=0 along x-axis. The position of the particle at an instant is given by x= ( 5 cm) sin pi t . Find the average velocity and average acceleration for a time interval 0-0.5 s . Hint : Average velocity =( x_(2)- x_(1))/(t_(2)-t_(1)) x_(1) = 5 sin 0 x_(2) = 5 sin ( pi xx 0.5) t_(2) - t_(1) = 0.5 Average acceleration = ( v_(2) - v_(1))/( t_(2) - t_(1)) v=(dx)/(dt) = 5 pi cos pit implies a _(av) = ( 5 pi cos ( pi xx 0.5) -5 pi cos theta)/(0.5)

A particle is moving with a position vector, vec(r)=[a_(0) sin (2pi t) hat(i)+a_(0) cos (2pi t) hat(j)] . find Distance travelled by the particle in 1 sec is

The velocity of a particle is u=v_0 + gt + ft^2 . If its position is x = 0 at t=0, Then it's displacement after time (t = 1 ) is

A particle is subjected to two simple harmonic motions given by x_(1) = 2.0sin (100 pi t) and x_(2) = 2.0sin (120pi t + pi //3) where, x is in cm and t in second. Find the displacement of the particle at (a) t = 0.0125 , (b) t = 0.025 .

A particle is subjected to two simple harmonic motions given by x_(1) = 2.0sin (100 pi t) and x_(2) = 2.0sin (120pi t + pi //3) where, x is in cm and t in second. Find the displacement of the particle at (a) t = 0.0125 , (b) t = 0.025 .

The displacement of a body executing SHM is given by x=A sin (2pi t+pi//3) . The first time from t=0 when the velocity is maximum is

The velocity of a particle is given by v=u_(0) + g t+ 1/2 at^(2) . If its position is x =0 at t=0 , then what is its displacement after t=1 s ?