ƒ(x) said:
Homework Statement
A truck travel 400 meters north in 80 seconds, and then it traveled 300 meters east in 70 seconds. The magnitude of the average velocity of the truck is most nearly:
a) 1.2 m/s
b) 3.3 m/s
c) 4.6 m/s
d) 6.6 m/s
e) 9.3 m/sHomework Equations
\bar{v} = Δx/Δt
The Attempt at a Solution
Δx = 500 m
Δt = 150 s
vav = 500/150 = 3.3 m/s
But, when I asked my teacher, she said:
vav = (300/70)i + (400/80)j = (30/7)i + 5j
||vav|| = 6.6 m/s
Why would you do it the second way and not the first?
I'm with the guy above me. I think your teacher is wrong.
During the first trip from 0 to 400, displacement is a simple function of distance, so it can be given by:
R_{80}(t) = \frac{R_f-R_i}{T_{total}}t
R_{80}(t) = \frac{400}{80}t
Therefore, the velocity for this function for the first eighty seconds is R(t) differentiated:
V_{80}(t) = \frac{400}{80}=5
At t = 80, displacement becomes a function of something else relative to the origin, so we can write this for t = 80 to 150
R_{70}(t) = \frac{R_f-R_i}{T_{total}}t
R_{70}(t) = \frac{\sqrt{400^2 + 300^2}-400}{70}t
velocity becomes
V_{70}(t) = \frac{\sqrt{400^2 + 300^2}-400}{70}=1.43
We then apply the calculus definition to average a function by integrating over some x and dividing by it:
V_{avg} = \frac{\int_{0}^{80} 5\, dy+\int_{0}^{70} 1.43\, dx}{70+80}=3.33
I took this overworking approach in hopes that I would arrive at your teacher's answer (so I could explain why it works), but I arrived at your method.
