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若把从西北路到北门的路程依次等分为a、b、c三段，一同学骑车从西北路到到北门时，在a段运行的平均速度是8m/s；在b段运行的时间比a段少∆t，b段的平均速度是9...
Mar 19, 2024
若把从西北路到北门的路程依次等分为a、b、c三段，一同学骑车从西北路到到北门时，在a段运行的平均速度是8m/s；在b段运行的时间比a段少∆t，b段的平均速度是9m/s；c段运行的时间比a段多∆t，则该火车在c段运行的平均速度是（ ） A．7m/s B．7.2m/s C．7.5m/s D．8.5m/s
Solution
a
Let the lengths of segments a, b, and c be La L_a , Lb L_b , and Lc L_c respectively. Since the segments are equal, La=Lb=Lc L_a = L_b = L_c
b
The time taken to travel segment a is ta=Lava t_a = \frac{L_a}{v_a} , where va=8 m/s v_a = 8 \, m/s
c
The time taken to travel segment b is tb=ta−Δt t_b = t_a - \Delta t , and the average speed for segment b is vb=9 m/s v_b = 9 \, m/s
d
The time taken to travel segment c is tc=ta+Δt t_c = t_a + \Delta t
e
The distance traveled in segment b is Lb=vb⋅tb=9(ta−Δt) L_b = v_b \cdot t_b = 9(t_a - \Delta t) . Since La=Lb L_a = L_b , we have 8ta=9(ta−Δt) 8t_a = 9(t_a - \Delta t)
f
Solve for Δt \Delta t to find Δt=ta−89ta=19ta \Delta t = t_a - \frac{8}{9}t_a = \frac{1}{9}t_a
g
Now, calculate tc=ta+Δt=ta+19ta=109ta t_c = t_a + \Delta t = t_a + \frac{1}{9}t_a = \frac{10}{9}t_a
h
The average speed for segment c is vc=Lctc=La109ta=8ta109ta=8⋅910=7.2 m/s v_c = \frac{L_c}{t_c} = \frac{L_a}{\frac{10}{9}t_a} = \frac{8t_a}{\frac{10}{9}t_a} = \frac{8 \cdot 9}{10} = 7.2 \, m/s
B
Key Concept
In kinematics, the average velocity is the total displacement divided by the total time taken. When segments of a trip are equal in length, the average velocities can be related to the times taken to traverse each segment.
Explanation
By setting up equations for the distances traveled in each segment and using the given average speeds and time differences, we can solve for the unknown average speed in segment c.
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