Question 1109988
{{{s(t) = 2.5t + 18}}} , with the understanding that
{{{s}}}= position of a model train, defined as distance in feet measured along the track from some point stated reference point, in a chosen direction,
and
{{{t}}}= time in seconds after an agreed upon {{{t=0}}} moment in time.
Maybe the there are milepost signs along the track, at 1 ft intervals, with milepost 0, at the train station marks position {{{s=0}}} .
Maybe time is measured from the moment the train start moving, and that time is taken as time {{{t=0}}} .
Anyway, at the time {{{t=0}}} , the train position is {{{s=18}}} ,
18 ft along the track (18 ft away in the chosen direction from the {{{s=0}}} position).
 
(a) At time {{{t=0}}} , the train is at position {{{s(0) = 2.5*0 + 18=18}}}
One second later, at time {{{t=1}}} ,
the train is at position {{{s(1) = 2.5*1 + 18=2.5+18}}} ,
{{{2.5 feet}}} further along the track going in the chosen direction.
During that {{{1second}}} , the train moved {{{2.5 feet}}} ,
so the model train is moving at an average speed of {{{2.5feet/"1 second"}}}{{{"="}}}{{{highlight("2.5 ft / second")}}} .
Obviously, in the next second it will be another {{{2.5feet}}} further along,
and we can say the average speed during that second is the same .
It turns out that the average speed is the same for every time period considered,
no matter when or for how long we measure the model train's speed.
That it is easy to see and understand.
We just make it sound complicated with fancy mathematical formulas and fancy words.
The physics teacher will say the model train is moving at constant speed,
and that {{{"2.5 ft / second"}}} is the "instant velocity" of the train at all times.
The math teacher will say that {{{s(t)}}} is a linear function,
and that the rate of change of the function with {{{t}}} is {{{2.5}}} .
You would be told that the graph of {{{s}}} versus {{{t}}}
is a straight line with slope {{{m=2.5}}}
(they like {{{m}}} as name for the slope).
The calculus teacher will say the derivative of function {{{s(t)}}} with respect to variable {{{t}}} is
{{{ds/dt=2.5}}} .
 
(b)  At {{{t=4}}} , 4 seconds after the time we agree to call {{{t=0}}} ,
the value of the function is
{{{s(4)=2.5*4+18}}}{{{"="}}}{{{10+18}}}}}}{{{"="}}}{{{28}}} ,
and the model train is at position {{{highlight(s=28)}}} .
{{{highlight(28feet)}}} from the {{{s=0}}} reference point, in the chosen direction.
 
(c) The train will be at position {{{s=33}}} when {{{s(t)=33}}} ,
or {{{2.5t+18=33}}} .
We have to find the time {{{t}}} by solving that equation.
{{{2.5t+18=33}}}
{{{2.5t=33-18}}}
{{{2.5t=15}}}
{{{t=15/2.5}}}
{{{highlight(t=6)}}} .
The train will be 33 ft along the track after {{{highlight(6seconds)}}} .