Recent content by figuringphysics

I see, will remember to include units in the future. The change to the second equation is essentially shifting it to the right by 1s to maintain continuity if I'm understanding correctly - is it valid to do this in this situation? Related to my reply to @WWGD. And following from this, would...

Thank you. I'm just not sure how we know to integrate from t=0s for the second equation, how it is possible to "reset" the time like this for the second equation?

The velocity is 3 m/s according to the first equation, and 3/e m/s according to the second: We would "solve" this discontinuity by adding a term representing the initial velocity (##v(x)=3e^{-t} + \frac{3e-3}{e}##), but then integrating gives: $$x(t)=-3e^{-t} + \frac{3e-3}{e}t + C_1$$ which...

My answer is d = (e+3)/e x(t) = ∫01 3t2 dt (0 ≤ t ≤1) = t3 |01 = 13 - 03 = 1m x(t) = ∫1∞ 3e-t dt (t > 1) = -3e-t |1∞ = lim(t->∞)[-3e-t] - [-3e-1] = 0 + 3e-1 = 3/e m Therefore total distance = 1m + 3/e m = (e+3)/e m However, the textbook answer gives...