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Question

A particle moves along x-axis and its acceleration at any time t is a=2sin(πt), where t is in seconds and a is in m/s2. The initial velocity of particle (at time t=0) is u=0.
Then the magnitude of displacement (in meters) by the particle from time t=0 to t=t will be
  1. 2π2sinπt2tπ
  2. 2π2sinπt+2tπ
  3. 2tπ
  4. None of these


Solution

The correct option is B 2π2sinπt+2tπ
a=sinπt dv=2sinπtdt
v=2πcosπt+C
At t=0, v=0 (u=0)
C=2π
v=2π(1cosπt)
Note: Velocity is always non-negative as cosθ1. Hence, particle always moves along positive x - direction.
Since velocity, v=dSdt
  Distance from time t=0 to t, S=t02π(1cosπt)dt=2π[t1πsinπt]t0=2πt2π2sinπt
Since the direction of velocity of particle is not changing, we can say
Displacement of the particle = Distance covered by the particle
Displacement from time t=0 to t s=2tπ2π2sinπt

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