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In the figure given, the position-time g...

In the figure given, the position-time graph of a particle of mass 0.1 kg is shown. The impulse at t=2 sec is
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Text Solution

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`v=(dx)/(dt)=4/2 = 2m//s `
velocity between t=0 sec to t =2 sec = 2 m/s
velocity at 2 sec = 0 m/s
Impulse = change in linear momentum
`=m(V_f - V_i) =0.1(0-2) = -0.2 kg ms^(-1)`
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