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Laws of Motion question

2018 · Shift 2 · Q41
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Laws of Motion question

2018 · Shift 2 · Q41

JEE AdvancedPhysicsLaws of MotionNumerical+3 / −1
A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m=0.4kgm=0.4kgm=0.4kg is at rest on this surface. An impulse of 1.0Ns1.0Ns1.0Ns is applied to the block at time t=0t=0t=0 so that it starts moving along the xxx-axis with a velocity v(t)=v0e−t/τ,v\left( t \right) = {v_0}{e^{ - t/\tau }},v(t)=v0​e−t/τ, where v0{v_0}v0​ is a constant and τ=4s.\tau = 4s.τ=4s. The displacement of the block, in metres, at t=τt = \taut=τ is ‾\underline{\hspace{2cm}}​ Take e−1=0.37.{e^{ - 1}} = 0.37.e−1=0.37.
Numerical answer
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