Calculus: The Language of Change

Exercise Set 10.1
Acceleration

by John Robeson

Exercise 1) 9.2.1 Again

The first exercise seeks your everyday interpretation of the positive and negative signs of (ds/dt)and (((d^2) s)/(dt^2)) on the hypothetical trip from 
Exercise 9.2.1.  We need to understand the mechanical interpretation of these derivatives as well as their graphical interpretation.

Look up your old solution to Exercise 9.2.1 and add a graphing table like the ones from Chapter 9 with slope and bending.  Fill in the parts of the table corresponding to (ds/dt) and (((d^2) s)/(dt^2)) using the microscopic slope and smile and frown icons including + and - signs.  Remember that (((d^2) s)/(dt^2)) is the derivative of the function (ds/dt); so, for example, when it is positive, the function v[t] = (ds/dt) increases, and when it is negative, the velocity decreases.  We also need to connect the sign of  (((d^2) s)/(dt^2))  with physics and the graph of s[t].

(a)  Where is your speed increasing?  Decreasing?  Zero?  If speed is increasing, what geometric shape must that portion of the graph of s[t] have?  (The graph of v[t] has upward slope and positive derivative, (dv/dt) = (((d^2) s)/(dt^2))  ≻ 0, but we are asking how an increase in v[t]  = (ds/dt) affects the graph of s[t].)

(b)  Is (ds/dt) ever negative in your example?  Could it be negative on someone's solution?  Why does this mean that you are backing up?

(c)  Summarize both the mechanics and geometrical meaning of the sign of the second derivative (((d^2) s)/(dt^2)) in a few words.  When (((d^2) s)/(dt^2)) is positive. . .    When (((d^2) s)/(dt^2))  is negative. . .

(d)  Why must  (((d^2) s)/(dt^2)) be negative somewhere on everyone's solution?

Solution

Here is one solution.

t(min)	0	2	8	10	10.5	12.5	24.5	26	104	105.5	111.5	113.5	114	120

s(mi.)	0	0.5	3	3.5	3.5	4	9	10	94.5	95.5	98	98.5	98.5	100

s'	     ↗	     ↗	     ↗	     →	     ↗	     ↗	     ↗	   ↗	     ↗	     ↗	     ↗	     →	     ↗

s''	     ∨	     *	     ∧	     *	     ∨	     *	     ∨	    *	     ∧	     *	     ∧	     *	    ∨∧

(a)  Where is your speed increasing?  Decreasing?  Zero?  If speed is increasing, what geometric shape must that portion of the graph of s[t] have?  (The graph of v[t] has upward slope and positive derivative, (dv/dt) = (((d^2) s)/(dt^2))  ≻ 0, but we are asking how an increase in v[t]  = (ds/dt) affects the graph of s[t].)

Solution to (a)

Our speed is increasing in segments 1, 5, 7, 13 which is whenever s'' has a  smile entry.  It is decreasing in segments 3, 9, 11, 13 which is when s'' has a frown entry.  The speed is zero when we are stopped during segments 4 and 12.

So when speed is increasing our graph of s[t] is smile shaped.

(b)  Is (ds/dt) ever negative in your example?  Could it be negative on someone's solution?  Why does this mean that you are backing up?

Solution to (b)

In the above solution(ds/dt)is never negative, but it could be if someones's solution contained backward travel.   If (ds/dt) were negative, then the change in distance would be negative, which means we are moving closer back to home.

(c)  Summarize both the mechanics and geometrical meaning of the sign of the second derivative (((d^2) s)/(dt^2)) in a few words.  When (((d^2) s)/(dt^2)) is positive. . .    When (((d^2) s)/(dt^2))  is negative. . .

Solution to (c)

When (((d^2) s)/(dt^2)) is positive the reading on your speedometer is moving upward and the slope of the graph is increasing.

When (((d^2) s)/(dt^2))  is negative the reading on your speedometer is moving downward and the slope of the graph is decreasing

(d)  Why must  (((d^2) s)/(dt^2)) be negative somewhere on everyone's solution?

Solution to (d)

(((d^2) s)/(dt^2)) must be negative somewhere because we must slow down at some point.  In other words our speedometer must move downward sometime or we would never stop.

Exercise 2) Numerical Acceleration

Compute the average speeds corresponding to the positions in Table 10.4 and write them next to the correct midpoint times so that they correspond to continuous velocities at those times.  Then use your velocities to compute accelerations at the proper times.  Simply fill in the places where the question marks appear in the velocity and acceleration Tables 10,5 and 10.6.  The data are also contained in the Gravity program so you can complete this arithmetic with the computer in Exercise 10.2.2.

Solution

The four unknown boxes in Table 10.5 are:

\(\(\((11.03 - 4.901)\)/\((1.5 - 1)\)
\)\)

\(\(\((19.6 - 11.03)\)/\((2 - 1.5)\)
\)\)

\(\(\((30.63 - 19.6)\)/\((2.5 - 2)\)
\)\)

\(\(\((44.1 - 30.63)\)/\((3 - 2.5)\)
\)\)

The four unknown boxes in Table 10.5 are:

\(\(\((12.258 - 7.356)\)/\((1.25 -  .75)\)
\)\)

\(\(\((17.14 - 12.258)\)/\((1.75 - 1.25)\)
\)\)

\(\(\((22.06 - 17.14)\)/\((2.25 - 1.75)\)
\)\)

\(\(\((26.94 - 22.06)\)/\((2.75 - 2.25)\)
\)\)

Observe that the acceleration is essentially constant at 9.8.