- #1

figuringphysics

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- Homework Statement
- A particle travels at velocity v(t)=3t^2 for 1s and then at v(t) = 3e^-t forever. How far does it travel?

- Relevant Equations
- x(t) = ∫ v(t) dt

My answer is d = (e+3)/e

x(t) = ∫

= t

= 1

= 1m

x(t) = ∫

= -3e

= lim(t->∞)[-3e

= 0 + 3e

= 3/e m

Therefore total distance = 1m + 3/e m =

However, the textbook answer gives

x(t) = ∫

= -3e

= 3 m

But I don't think this is correct. Am I missing something in the question? Thanks

x(t) = ∫

_{0}^{1}3t^{2}dt (0 ≤ t ≤1)= t

^{3}|_{0}^{1}= 1

^{3}- 0^{3}= 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

**4m**. I can see how this is possible if we integrate the second equation from 0 to infinity:x(t) = ∫

_{0}^{∞}v(t) dt (1 < t)= -3e

^{-t}|_{0}^{∞}= 3 m

But I don't think this is correct. Am I missing something in the question? Thanks