A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 W. The remainder of the day she is sitting in class, studying, or resting; during these activities, she expends energy at an average rate of 100 W. If she expends a total of $\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{\hspace{0.33em}}\mathit{J}$ of energy in a 24 hour day, how much of the day did she spend walking?

It is known that 24 h = 86400 s.

If student takes$t_1$ seconds to walk, the time taken for student to sit is $\left(86400-t_{{}_1}\right)s$.

Total energy expanded is $E=1.1\times {10}^7\hspace{0.33em}J$.

Amount of power expanded by student to walk is $p_1=280\hspace{0.33em}W$.

Amount of power expanded by student to sit is $p_2=100\hspace{0.33em}W$.

The relationship between the relationship between the power, energy, and time is given by,

$\mathit{E}\mathbf{=}\mathit{P}\mathit{t}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, P, is power, E is energy, and t is time.

The total time of the whole day is,

$$\begin{gathered} t=t_{walk}+t_{other} \\ =86400...........\left(2\right) \end{gathered}$$

The time spend for other activities will be as follows.

role="math" localid="1663759984149" $t_{other}=86400-t_{walk}...........\left(3\right)$

From equation (1),

$$\begin{gathered} E=P_{walk}{t}_{walk}+P_{other}{t}_{other} \\ =P_{walk}{t}_{walk}+P_{other}\left(86400-t_{walk}\right) \\ =\left(P_{walk}{t}_{other}\right)t_{walk}+P_{other}\left(86400\right) \\ {t}_{walk}=\frac{E-P_{other}\left(86400\right)}{\left(P_{walk}{P}_{other}\right)}...........\left(4\right) \end{gathered}$$

Substitute all the values in equation (4).

$$\begin{gathered} {\mathrm{t}}_{\mathrm{walk}}=\frac{1.1\times {10}^7\mathrm{J}-100\mathrm{W}\times 86400}{\left(280\mathrm{W}-100\mathrm{W}\right)} \\ =\frac{236\times {10}^4}{180}\mathrm{s} \\ =\left(1.31\times {10}^4\mathrm{s}\right)\left(\frac{1\mathrm{hr}}{60\times 60\mathrm{s}}\right) \\ =3.64\mathrm{hr} \end{gathered}$$

Therefore, the required time is 3.64 hr.