A car moves along an x axis through a distanceof 890 m, starting at rest(at x =

Question

A car moves along an x axis through a distanceof 890 m, starting at rest(at x = 0) and ending at rest(at x = 890 m). Through the first 1/4 of that distance,its acceleration is +2.55 m/s2. Through the next 3/4 ofthat distance, its acceleration is -0.850 m/s2. (a) What is its travel time through the 890 m?
s

(b) What is its maximum speed?
m/s

(c) Graph position x, velocity v, andacceleration a versustime t for the trip. (Do this on paper. Yourinstructor may ask you to turn in this work.)
(a) What is its travel time through the 890 m?
s

(b) What is its maximum speed?
m/s

(c) Graph position x, velocity v, andacceleration a versustime t for the trip. (Do this on paper. Yourinstructor may ask you to turn in this work.)

Explanation / Answer

he actually forget about t2 which is the time needed to go throughthe second part of the distance. t2=v/a2=33,6/0,85=39,5(s) So the total time is 39,5+13,21=52,7(s)

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