Rate of Change with Derivatives We just started with rate of change while using derivatives and I am stuck on a question, hope you can help. A particle moves on a vertical line so that its altitude at time t is y=t^3−12*t+3, where t≥0. Find the distance that the particle travels in the time interval 0<=t<=3.

Sydney Stein

Sydney Stein

Answered question

2022-08-10

Rate of Change with DerivativesWe just started with rate of change while using derivatives and I am stuck on a question, hope you can help.
A particle moves on a vertical line so that its altitude at time t is y = t 3 12 · t + 3, where t 0. Find the distance that the particle travels in the time interval 0 t 3

Answer & Explanation

Izabella Fisher

Izabella Fisher

Beginner2022-08-11Added 14 answers

If the particle's position function was monotonic, then this question would be easy; all you would have to do is compute the net displacement | y ( 3 ) y ( 0 ) | . Unfortunately, the particle changes direction. To find this turning point, compute the derivative and solve for the time value t 0 ( 0 , 3 ) where y ( t 0 ) = 0. Then the distance will be given by:
| y ( 3 ) y ( t 0 ) | + | y ( t 0 ) y ( 0 ) |

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