Respuesta :
Given,
Time duration to reach the 1st city, t₁=55.0 min
The speed at which the driver drove to 1st city, s₁=65.0 km/h
Time duration to reach the 2nd city, t₂=16.0 min
The speed at which the driver drove to 2nd city, s₂=70 km/hr
Time duration to reach the 3rd city, t₃=60.0 min
The speed at which the driver drove to 3rd city, s₃=45.0 km/hr
The time spent for lunch and buying gas, t₄=20.0 min
The speed of an object is given by,
[tex]s=\frac{d}{t}[/tex]Thus the distance traveled by the object is given by,
[tex]d=st[/tex]The distance between the 1st pairs of the city is given by
[tex]d_1=s_1t_1[/tex]On substituting the known values,
[tex]\begin{gathered} d_1=65\times\frac{55}{60} \\ =59.58\text{ km} \end{gathered}[/tex]Distance between the second pair of the cities is,
[tex]\begin{gathered} d_2=s_2t_2 \\ =70\times\frac{16}{60} \\ =18.67\text{ km} \end{gathered}[/tex]The distance between the 3rd pair of cities is
[tex]\begin{gathered} d_3=s_3t_3 \\ =45\times\frac{60}{60} \\ =45\text{ km} \end{gathered}[/tex](a) The average speed is defined as the total distance divided by the total time taken.
Therefore the average speed of the whole trip is calculated as
[tex]s_{av}=\frac{d_1+d_2+d_3}{t_1+t_2+t_3+t_4}[/tex]On substituting the known values in the above equation,
[tex]\begin{gathered} s_{av}=\frac{59.58+18.67+45}{\frac{(55+16+60+20)}{60}} \\ =\frac{123.25}{2.52} \\ =48.91\text{ km/hr} \end{gathered}[/tex]Therefore the average speed of the trip is 48.91 km
(b) The distance between the initial and the final cities is given by,
[tex]D=d_1+d_2+d_3[/tex]On substituting the known values in the above equation,
[tex]\begin{gathered} D=59.58+18.67+45 \\ =123.25\text{ km} \end{gathered}[/tex]Therefore the total distance between the initial and the final cities is 123.25 km
