The distabce travelled in a time, t by a particle moving along `x`-axis according to an equation `x = A cos omegat` is given by `S = nA +S_(0)`, when time `t` can be written as `t =n((T)/(4)) +t_(0)` such that `t_(0) lt T//4` and distance travelled in that `t_(0)` is `S_(0)`. The distance `S` is, when 'n' is even number:
A
`S = A [(n+1)-cos (omegat +(npi)/(2))]`
B
`S = A[(n+1)-cos (omegat-(npi)/(2))]`
C
`S =A [A +sin (omegat -(npi)/(2))]`
D
`S = A [(n+1)-sin (omegat -(npi)/(2))]`
Text Solution
Verified by Experts
The correct Answer is:
B
`S = nA +S_(0) , S_(0) = A - x_(0) , x_(0) = A cos wt_(0)`
The distabce travelled in a time, t by a particle moving along x -axis according to an equation x = A cos omegat is given by S = nA +S_(0) , when time t can be written as t =n((T)/(4)) +t_(0) such that t_(0) lt T//4 and distance travelled in that t_(0) is S_(0) . The distance S is, when 'n' is odd number.
The distance travelled (in meters) by the particle from time to t = 0 to t = 1s will be –
The distance travelled (in meters) by the particle from time t = 0 to t = t will be –
If the distance s travelled by a particle in time t is s=a sin t +b cos 2t , then the acceleration at t=0 is
The distance s travelled by a particle moving on a straight line in time t sec is given by s=2t^(3)-9t^(2)+12t+6 then the initial velocity of the particle is
A particle moves along the x axis according to the law x=a cos omega t . Find the distance that the particle covers during the time interval from t=0 to t .
A particle is moving along the x-axis whose acceleration is given by a= 3x-4 , where x is the location of the particle. At t = 0, the particle is at rest at x = 4//3m . The distance travelled by the particles in 5 s is
The distance travelled s in metres by a particle in t second is given by s=t^(3)+2t^(2)+t . The speed of the particle after 1s will be
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 ?
If S=2t^(3)-4t^(2)+12t , then distance travelled during the time -interval [0,1] is
Knowledge Check
The distabce travelled in a time, t by a particle moving along x -axis according to an equation x = A cos omegat is given by S = nA +S_(0) , when time t can be written as t =n((T)/(4)) +t_(0) such that t_(0) lt T//4 and distance travelled in that t_(0) is S_(0) . The distance S is, when 'n' is odd number.
A
`S =A [(n+1) -cos (omegat+(npi)/(2))]`
B
`S = A[(n+1)-cos (omegat-(npi)/(2))]`
C
`S = A[n +sin (omegat -(npi)/(2))]`
D
`S = A [(n+1)-sin (omegat-(npi)/(2))]`
If the distance s travelled by a particle in time t is s=a sin t +b cos 2t , then the acceleration at t=0 is
A
`a`
B
`-a`
C
`4b`
D
`-4b`
The distance s travelled by a particle moving on a straight line in time t sec is given by s=2t^(3)-9t^(2)+12t+6 then the initial velocity of the particle is
