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)
a 18 9
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.

Step2: Calculate velocity for leg A

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

Step3: Calculate velocity for leg B

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

Step4: Calculate velocity for leg C

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

Step5: Calculate velocity for leg D

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

Step6: Calculate velocity for leg E

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

Step7: Arrange velocities from highest to lowest

$v_D = 240\ km/h$, $v_C = 180\ km/h$, $v_E\approx128.57\ km/h$, $v_A = 120\ km/h$, $v_B = 100\ km/h$
So the order is D, C, E, A, B

Answer:

D, C, E, A, B