Acar moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at
rest (at x = 900 m). through the first ¼ of that distance, its acceleration is +2.25 m/s2
. through the rest of that distance, its acceleration is – 0.750 m/s2
. find its maximum speed?

A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first 1/4 of that distance, its acceleration is +6.25 m/s2. Through the next 3/4 of that distance, its acceleration is -2.08 m/s2. What are (a) its travel time through the 900 m and (b) its maximum speed? 

Solve for the time at the 1/4 mark. That's 225 m. How? d = (1/2)at^2 ( initial velocity zero). Thus 225 = (1/2) 6.25 t^2. t^2 = ( 225 * 2 ) / 6.25. t = 8.5 sec. 

At the other end t^2 = (675 * 2) / 2.08 -- we reversed the sign and ran time backwards. t = 25.5 sec. 

So total time is 8.5 + 25.5 or 34 sec. 

Since zero initial velocity: v^2 = 2 a d. Here, v^2 = 2 * 6.25 * 225. v = 53 m/s. That's the fastest speed since braking then occurs.

Iread this book last year so take this with a grain of salt.gene jumps from the tree to compete with finny due to the fact that gene is always trying to do better than finny. he is put under the grasp of peer pressure due to finny almost forcing him to do this. ever since he first jumped finny kept pressuring him to jump since he thought gene enjoyed jumping.rather than uses this answer you should probably just reread chapters 3-4