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.

1.state2.trash3.the earth

Answer the above statement is ( false );