It is important to notice firs of all that the train with no stop acceleration and them no stop acceleration would have a faster travel between stops. given this is safe to assume that there is a period during while the train is not accelerating nor decelerating. To find the time that is speed is constant
let's assume non-stop accelerating and decelerating and the time lost (subtraction of both) is the time the train had constant speed:
assuming same time accelerating a decelerating we have that the distance is
x = voT + 1/2aT^2 (initiial speed = 0) (0.25 miles = 1320 feets)
1320 = 1/2 (8)(T/2)^2
T/2 = 18.17 seg
T = 36.34 seg
time of constant speed
T = 41 -36.34 = 4.66 seg
maximum speed
v = v0 + aT (initial speed = 0)
v = 8(18.17) (time accelerating)
v = 145.36 feets/sec
distance at full speed
x = v*t
x = 145.36*4.66 = 677.38 feets
it's d just put them together to equal y
Answer:
-1/4
Step-by-step explanation:
Multiplying 64 by -1/4 results in -16;
mult. -16 by -1/4 results in 4,
and so on.
The common ratio is -1/4.
Should be the first option.
Since ABP is congruent to CBP in the 4th step, AP must be equal to CP
Answer:
1.38 ft
Step-by-step explanation:
42 cm x <u> 3.28 ft </u>
100 cm
= 1.38 ft
there are plenty of printable unit conversion on google
