Answer:
The answer should be 1500g I think.
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
Answer:
2+2= 4
Step-by-step explanation:
just have a scenario like : you have two apples and your friend brings two more apples when you count them all, there are four apple
Answer:
y = 2x +13
Step-by-step explanation:
when 2 lines are perpendicular, the product of the two gradients is -1
i.e. m x m1 = -1
so from 10y + 5x = 3, the gradient m is -0.5
m1 = -1 / -0.5 = 2
using m1 = 2 and pt(-7,-1),
the eqn is y = (m1)x +c
-1 = 2(-7) + c
c= 13
thus y = 2x +13
