The position of a car driving along a straight road at time $t$ in minutes is given by the function y = s(t) that is pictured. The car's position function has units measured in thousands of feet. Remember that you worked with this function and sketched graphs of y = v(t) = s'(t) and y = v'(t)

The graph of y = s(t), the position of the car (measured in thousands of feet from its starting location) at time t in minutes.

(a) On what intervals is the position function $y = s(t)$ increasing? decreasing? Why?

(b) On which intervals is the velocity function $y = v(t) = s'(t)$ increasing? decreasing? neither? Why?

(c) Acceleration is defined to be the instantaneous rate of change of velocity, as the acceleration of an object measures the rate at which the velocity of the object is changing. Say that the car's acceleration function is named a(t). How is a(t) computed from v(t)? How is a(t) computed from s(t)? Explain.

(d) What can you say about s'' whenever s' is increasing? Why?

(e) Using only the words \(

*increasing<\i>\) \(**decreasing<\i>\), \(**constant<\i>\), \(**concave up<\i>\), \(**concave down<\i>\), and \(**linear<\i>\), complete the following sentences. For the position function s with velocity v and acceleration a,*

\(\bullet\) on an interval where v is positive, s is \(\underline{\hspace{1.5in}}\).

\(\bullet\) on an interval where v is negative, s is \(\underline{\hspace{1.5in}}\).

\(\bullet\) on an interval where v is zero, s is \(\underline{\hspace{1.5in}}\).

\(\bullet\) on an interval where a is positive, v is \(\underline{\hspace{1.5in}}\).

\(\bullet\) on an interval where a is negative, v is \(\underline{\hspace{1.5in}}\).

\(\bullet\) on an interval where a is zero, v is \(\underline{\hspace{1.5in}}\).

\(\bullet\) on an interval where a is positive, s is \(\underline{\hspace{1.5in}}\).

\(\bullet\) on an interval where a is negative, s is \(\underline{\hspace{1.5in}}\).

\(\bullet\) on an interval where a is zero, s is \(\underline{\hspace{1.5in}}\).