Question

adam takes a bus on a school field trip. the bus route is split into the five legs listed in the table. find the average velocity for each leg of the trip. then arrange the legs of the trip from highest velocity to lowest.

leg	distance (km)	time (min)
b	25	15
c	24	8
d	48	12
e	15	7

Explanation:

Step1: Recall velocity formula

The formula for average velocity $v=\frac{d}{t}$, where $d$ is distance and $t$ is time. We need to convert time from minutes to hours since velocity is usually in km/h. 1 hour = 60 minutes, so $t$ (in hours)=$\frac{t_{min}}{60}$.

Step2: Calculate velocity for leg A

$t_A=\frac{9}{60}=0.15$ h, $d_A = 18$ km. Then $v_A=\frac{d_A}{t_A}=\frac{18}{0.15}=120$ km/h.

Step3: Calculate velocity for leg B

$t_B=\frac{15}{60}=0.25$ h, $d_B = 25$ km. Then $v_B=\frac{d_B}{t_B}=\frac{25}{0.25}=100$ km/h.

Step4: Calculate velocity for leg C

$t_C=\frac{8}{60}=\frac{2}{15}$ h, $d_C = 24$ km. Then $v_C=\frac{d_C}{t_C}=\frac{24}{\frac{2}{15}}=24\times\frac{15}{2}=180$ km/h.

Step5: Calculate velocity for leg D

$t_D=\frac{12}{60}=0.2$ h, $d_D = 48$ km. Then $v_D=\frac{d_D}{t_D}=\frac{48}{0.2}=240$ km/h.

Step6: Calculate velocity for leg E

$t_E=\frac{7}{60}$ h, $d_E = 15$ km. Then $v_E=\frac{d_E}{t_E}=\frac{15}{\frac{7}{60}}=\frac{900}{7}\approx128.57$ km/h.

Step7: Arrange velocities

Arranging from highest to lowest: $v_D > v_C>v_E > v_A>v_B$.

Answer:

Leg D: 240 km/h, Leg C: 180 km/h, Leg E: $\frac{900}{7}\approx128.57$ km/h, Leg A: 120 km/h, Leg B: 100 km/h