Question

The displacement x of particle moving in one dimension, under the action of a constant force is related to the time t by the equation t = sqrt(x) +3 where x is in meters and t in seconds . Find (i) The displacement of the particle when its velocity is zero , and (ii) The work done by the force in the first 6 seconds.

Answer 1

The displacement of the particle when its velocity is zero is 0

The work done by the force in the first 6 seconds is 18m.

• Given t = √x + 3

• x = ( t - 3)²

Therefore velocity,

• dx/dt = V = 2 ( t - 3 )

( i ) to find displacement of particle when velocity is 0 ,

• V = 0 ==> 2 ( t-  3 ) = 0
• t = 3
• x = (3 - 3 ) ²= 0
• displacement of particle when its velocity is 0 is 0.

(ii) to find the work done by F ,

• Work done  = F.X
• Since its one dimentional force acts along the direction of motion.
• Displacement in first 6 seconds = x = (6 - 3 ) ² = 9
• Work done = Fx9
• If mass of the particle is m Kg
• F = ma
• acceleration = d²x/dt² = dV/dt = 2m/s²
• F = 2m
• Therefore, work done = 18m .