A particle moves along the xxx-axis with acceleration a(t)=sin(t)+e−ta(t) = \sin(t) + e^{-t}a(t)=sin(t)+e−t. If v(0)=1v(0) = 1v(0)=1 and s(0)=0s(0) = 0s(0)=0, find s(t)s(t)s(t).
s(t)=2−sin(t)+e−ts(t) = 2 - \sin(t) + e^{-t}s(t)=2−sin(t)+e−t
s(t)=2t−sin(t)+e−t−1s(t) = 2t - \sin(t) + e^{-t} - 1s(t)=2t−sin(t)+e−t−1
s(t)=2t−sin(t)−e−t+1s(t) = 2t - \sin(t) - e^{-t} + 1s(t)=2t−sin(t)−e−t+1
s(t)=1+sin(t)−e−ts(t) = 1 + \sin(t) - e^{-t}s(t)=1+sin(t)−e−t