The position of a model train, in feet along a railroad track, is given bys(t) = 2.5t + 14after t seconds.(a) How fast is the train moving?ft/sec(b) Where is the train after 4 seconds?241 feet along the trackEnter an exact number.(c) When will the train be 29 feet along the track?t=sec

Answer:Part a) The speed is [tex]2.5\frac{ft}{sec}[/tex]Part b) After 4 seconds the trains is 24 ft along the trackPart c) [tex]t=6\ sec[/tex]Step-by-step explanation:we have[tex]s(t)=2.5t+14[/tex]This is the equation of a line in slope intercept formwheres(t) is the position of a model train in feett is the time in secondsPart a) How fast is the train moving?The speed of the train is equal to the slope of the linear equation soThe slope m is equal to[tex]m=2.5\frac{ft}{sec}[/tex]thereforeThe speed is [tex]2.5\frac{ft}{sec}[/tex]Part b) Where is the train after 4 seconds? For t=4 secsubstitute the value of t in the equation and solve for s[tex]s(4)=2.5(4)+14=24\ ft[/tex]thereforeAfter 4 seconds the trains is 24 ft along the trackPart c) When will the train be 29 feet along the track?For s(t)=29 ftSubstitute the value of s(t) in the equation and solve for t[tex]29=2.5t+14[/tex]subtract 14 both sides[tex]29-14=2.5t[/tex][tex]15=2.5t[/tex]Divide by 2.5 both sides[tex]6=t[/tex]rewrite[tex]t=6\ sec[/tex]