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The distance $s$, in metres, covered by a train on a straight track in the first two seconds is given by $s=\int_{0}^{2}20(1-e^{-t})dt$. Find $s$
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s =[20 t + 20e^-t] at T= 0 , s = 20 at T = 2 , s = 40+2.71= 42.71 S= 42.71 - 20 = 22.71m
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