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 be2π2sinπt−2tπ−2π2sinπt+2tπ2tπNone of these

Solution

## The correct option is B −2π2sinπt+2tπa=sinπt ∴∫dv=∫2sinπtdt v=−2πcosπt+C At t=0, v=0 (∵u=0) ∴C=2π ⇒v=2π(1−cosπt) Note: Velocity is always non-negative as cosθ≤1. Hence, particle always moves along positive x - direction. Since velocity, v=dSdt ∴  Distance from time t=0 to t, S=t∫02π(1−cosπt)dt=2π[t−1πsinπt]t0=2πt−2π2sinπt 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=2tπ−2π2sinπt

Suggest corrections