Question: Three digital clocks A, B, and C run at different rates and do not have simultaneous readings of zero. Figure 1-6 shows simultaneous readings on pairs of the clocks for four occasions. (At the earliest occasion, for example, B reads 25.0 s and C reads 92.0 s.) If two events are 600 s apart on clock A, how far apart are they on (a) clock B and (b) clock C? (c) When clock A reads 400 s, what does clock B read? (d) When clock C reads 15.0 s, what does clock B read? (Assume negative readings for prezero times.)

Figure 1-6Problem 13

The given quantities are as below:

Reading of clock B, t_b =25.0s

Reading of clock C, t_c =92.0s

Two events are 600 s away on clock A.

Write the equations defining the relationship between the times, on both scales, using the given information. Using these equations, solve it for the required time.

The time on any of these clocks is a straight-line function of that on another, with slopes 1 and y-intercepts. Use the equation of the straight line to calculate the above quantities.

Assume that,

$t_c=x{t}_b+y$ … (i)

Substitute the given values in equation (i).

$92=25x+y$ … (ii)

And,

$142=200x+y$ … (iii)

Solving equations (ii) and (iii) simultaneously for X and Y to get,

$x=\frac{2}{7},andy=\frac{594}{7}$

Substituting values of X and Y, write the equation (i) to define the relationship between t_b and t_c as,

$t_c=\frac{2}{7}{t}_b+\frac{594}{7}$ … (iv)

$$\begin{gathered} \text{Now,assumethat,} \\ {\text{t}}_b=P{t}_a+s\dots \left(v\right) \\ 125=312P+s\dots \left(vi\right) \\ 290=512P+s\dots \left(vii\right) \\ Solveequations\left(vi\right)and\left(vii\right)simultaneouslyforPandS, \\ P=\frac{33}{40} \\ S=-\frac{662}{5} \\ Substitutethevaluesinequation\left(v\right), \\ {t}_b=\frac{33}{40}{t}_a-\frac{662}{5} \\ Hence,therelationbetweenthetimesonclocksis \\ {t}_c=\frac{2}{7}{t}_b+\frac{594}{7}\dots \left(viii\right) \\ {t}_b=\frac{33}{40}{t}_a-\frac{662}{5}\dots \left(ix\right) \end{gathered}$$

Using equation (viii) and (ix),

$$\begin{gathered} t_b-t_b=\frac{33}{40}(t{\text{'}}_a-t_a) \\ =\frac{33}{40}\times \left(600\right)\sin ce(t_a-t_a=600s) \\ =495s \end{gathered}$$

Thus, reading on clock B is 495 s when two events on clock A are 600 s apart.

Now, find time on clock C.

$$\begin{gathered} t_c-t_c=\frac{2}{7}(t_b-t_b) \\ =\frac{2}{7}\times \left(495\right) \\ =141s \end{gathered}$$

Thus, reading on clock C is 141 when two events on clock A 6 00sareapart.

Find reading on clock B when clock A readss.

$$\begin{gathered} t_b=\frac{33}{40}\times \left(400\right)-\frac{662}{5} \\ =198s \end{gathered}$$

Thus, reading on B is 198 s.

Find reading on clock B when clock C reads.

$$\begin{gathered} t_c=\frac{2}{7}{t}_b+\frac{594}{7} \\ 15=\frac{2}{7}{t}_b+\frac{594}{7} \\ {t}_b=-245s \end{gathered}$$

Thus, reading on clock B is -245 s.