Question

a bus travels 27 km due east at an average speed of 18 m/s, making stops along the way. the bus then stops at the bus station for 55 minutes for a lunch break and then continues due east another 11 km for 7 minutes. what is the buss average velocity over this entire drive?

Explanation:

Step1: Convert units

• Convert distance: \(27\space km = 27\times10^{3}\space m\), \(11\space km=11\times 10^{3}\space m\)
• Convert time for the first - part: \(v = 18\space m/s\), \(d_1=27\times 10^{3}\space m\), using \(t=\frac{d}{v}\), \(t_1=\frac{27\times 10^{3}}{18}=1500\space s\)
• Convert time for the second - part: \(t_3 = 7\space min=7\times60 = 420\space s\)
• Convert time for the break: \(t_2=55\space min = 55\times60=3300\space s\)

Step2: Calculate total displacement and total time

• Total displacement \(d=d_1 + d_2=27\times10^{3}+11\times10^{3}=38\times10^{3}\space m\)
• Total time \(t=t_1 + t_2+t_3=1500 + 3300+420=5220\space s\)

Step3: Calculate average velocity

• Using the formula \(v_{avg}=\frac{d}{t}\), \(v_{avg}=\frac{38\times 10^{3}}{5220}\approx7.28\space m/s\)

Answer:

The bus's average velocity is approximately \(7.28\space m/s\) due East.