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(B) Given,$m = 0.4 \ kg$,Impulse $J = 1.0 \ N \ s$,$	au = 4 \ s$,and $v(t) = v_0 e^{-t/	au}$.At $t = 0$,the velocity is $v(0) = v_0 e^0 = v_0$.From

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$6.3$

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(B) Given,$m = 0.4 \ kg$,Impulse $J = 1.0 \ N \ s$,$	au = 4 \ s$,and $v(t) = v_0 e^{-t/	au}$.At $t = 0$,the velocity is $v(0) = v_0 e^0 = v_0$.From the impulse-momentum theorem,$J = \Delta p = m \Delta v = m(v(0) - 0) = m v_0$.Therefore,$v_0 = J/m = 1.0 / 0.4 = 2.5 \ m/s$.The displacement $S$ at $t = 	au$ is given by the integral of velocity:$S = \int_0^{	au} v(t) \ dt = \int_0^{	au} v_0 e^{-t/	au} \ dt$.$S = v_0 \left[ -	au e^{-t/	au} ight]_0^{	au} = v_0 	au (1 - e^{-1})$.Substituting the values: $S = (2.5) 	imes (4) 	imes (1 - 0.37)$.$S = 10 	imes 0.63 = 6.3 \ m$.

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