A force is applied to a body of mass 3 kg at | vedclass

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(D) The impulse applied to the body is equal to the area under the force-time $(F-t)$ graph.Impulse $(J)$ = Change in momentum $(\Delta p)$ = $m(v - u

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$2$

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(D) The impulse applied to the body is equal to the area under the force-time $(F-t)$ graph.Impulse $(J)$ = Change in momentum $(\Delta p)$ = $m(v - u)$.Given mass $m = 3 \ kg$, initial velocity $u = 0 \ m/s$.Area under the graph = Area of the first rectangle + Area of the second rectangle.Area = $(8 \ N 	imes 0.5 \ s) + (4 \ N 	imes (1.0 - 0.5) \ s)$.Area = $(4 \ Ns) + (4 \ N 	imes 0.5 \ s) = 4 \ Ns + 2 \ Ns = 6 \ Ns$.Since Impulse = $\Delta p = m(v - u)$,$6 = 3 	imes (v - 0)$,$6 = 3v$,$v = 2 \ m/s$.Therefore, the speed of the body after $1 \ s$ is $2 \ m/s$.

$A$ force is applied to a body of mass $3 \ kg$ at rest on a frictionless horizontal surface, as shown in the force-time $(F-t)$ graph. The speed of the body after $1 \ s$ is: (in $m/s$)

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