A baseball having mass of 0.4 kg is thrown s | vedclass

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(B) From the velocity-time graph,the acceleration $a = \frac{dv}{dt} = \frac{20}{0.3} = \frac{200}{3} \ m/s^2$. Since the mass $m = 0.4 \ kg$,the net

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The net force on the baseball is constant.

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(B) From the velocity-time graph,the acceleration $a = \frac{dv}{dt} = \frac{20}{0.3} = \frac{200}{3} \ m/s^2$. Since the mass $m = 0.4 \ kg$,the net force $F_{net} = ma = 0.4 	imes \frac{200}{3} = \frac{80}{3} \ N$,which is constant. Thus,option $B$ is incorrect because the force shown in the first graph is just one of the forces,not the net force.For $0 \le t \le 0.2 \ s$,$F(t) = 800 \ N$ and $v(t) = \frac{20}{0.3}t = \frac{200}{3}t$. Power $P = Fv = 800 	imes \frac{200}{3}t = \frac{160000}{3}t$,which is a linear function (straight line).For $0.2 \le t \le 0.3 \ s$,$F(t)$ is a linear function with slope $m = \frac{0 - 800}{0.3 - 0.2} = -8000$. So,$F(t) = 800 - 8000(t - 0.2) = 2400 - 8000t$. Power $P = Fv = (2400 - 8000t) 	imes \frac{200}{3}t = 160000t - \frac{1600000}{3}t^2$,which is a downward-opening parabola.

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