The position of a model train, in feet along a railroad track, is given by
s(t) = 2.5t + 14
after t seconds.
(a) how fast is the train moving?
ft/sec
(b) where is the train after 4 seconds?
241 feet along the track
enter an exact number.
(c) when will the train be 29 feet along the track?
t=
sec

Part a) The speed is

Part b) After 4 seconds the trains is 24 ft along the track

Part c)

Step-by-step explanation:

we have

This is the equation of a line in slope intercept form

where

s(t) is the position of a model train in feet

t is the time in seconds

Part a) How fast is the train moving?

The speed of the train is equal to the slope of the linear equation so

The slope m is equal to

therefore

The speed is

Part b) Where is the train after 4 seconds?

For t=4 sec

substitute the value of t in the equation and solve for s

therefore

After 4 seconds the trains is 24 ft along the track

Part c) When will the train be 29 feet along the track?

For s(t)=29 ft

Substitute the value of s(t) in the equation and solve for t

subtract 14 both sides

Divide by 2.5 both sides

rewrite

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.