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 is moving in a plane with a velocity given by, vecu=u_(0)hati+(omegacosomegat) hatj, are unit vectors along x and y-axes respectively. If the particle is at the origin an t=0, then its distance from the origin at time t=3pi//2omega will be

A particle is moving in a plane with velocity vec(v) = u_(0)hat(i) + k omega cos omega t hat(j) . If the particle is at origin at t = 0 , (a) determine the trajectory of the particle. (b) Find its distance from the origin at t = 3pi//2 omega .

A particle moves along the x - axis according to x = A[1 + sin omega t] . What distance does is travel in time interval from t = 0 to t = 2.5pi//omega ?

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)

Only one force is acting on a 2.8 kg particle-like object whose position is given by x= 4.0t-5.0t^(2)+2.0t^(3) , with x in meters and t in seconds. What is the work done by the force from t=0 to t= 6.0 s ?

A simple harmonic oscillator starts from mean position at time, t = 0, moves along +x-direction are reaches the extreme point in 1s covering a distance of 20 cm (a) The equation of its motion is x = 0.2 "cos" (pi)/(2) t (b) Its maximum speed is 0.314 ms^(-1) (c) Its angular frequency is (pi)/(2) rad s^(-1) (d) Its maximum acceleration is (pi^(2))/(100) ms^(-2)

int_(0)^(2 pi)[sin t]dt

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 .