Truck Average Velocity Calculation - 400m North, 300m East in 80s and 70s

[ƒ(x)]

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/s

Homework Equations

\bar{v} = Δx/Δt

The Attempt at a Solution

Δx = 500 m
Δt = 150 s

v_av = 500/150 = 3.3 m/s

But, when I asked my teacher, she said:

v_av = (300/70)i + (400/80)j = (30/7)i + 5j
||v_av|| = 6.6 m/s

Why would you do it the second way and not the first?

 

[tiny-tim]

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:
…
Δx = 500 m
Δt = 150 s

v_av = 500/150 = 3.3 m/s

But, when I asked my teacher, she said:

v_av = (300/70)i + (400/80)j = (30/7)i + 5j
||v_av|| = 6.6 m/s

Why would you do it the second way and not the first?

Hi ƒ(x)!

I'm with you …

average velocity = total (vector) displacement divided by total time = 500 north-east-ish / 150, so its magnitude is 500/150.

I don't understand your teacher's method at all.

 

[ƒ(x)]

The problem is that I can see the logic behind that method too...

Because there's a change in direction, you have to treat each part separately and break it into vector components...? yeah..but no, not really. That doesn't hold for the definition of average velocity...average speed yes, but not velocity, right? It would be much clearer if 6.6 wasn't one of the choices

 

[ƒ(x)]

can anyone clarify this for me?

 

[xcvxcvvc]

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/s

Homework Equations

\bar{v} = Δx/Δt

The Attempt at a Solution

Δx = 500 m
Δt = 150 s

v_av = 500/150 = 3.3 m/s

But, when I asked my teacher, she said:

v_av = (300/70)i + (400/80)j = (30/7)i + 5j
||v_av|| = 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.

 

[ƒ(x)]

Ok.

<br /> V_{avg} = \frac{\int_{0}^{80} 5\, dy+\int_{0}^{70} 1.43\, dx}{70+80}=3.33<br />

Why isn't it two separate fractions?

 

[xcvxcvvc]

Ok.

<br /> V_{avg} = \frac{\int_{0}^{80} 5\, dy+\int_{0}^{70} 1.43\, dx}{70+80}=3.33<br />

Why isn't it two separate fractions?

Because if you were to graph displacement with the starting position to be the origin of the graph, displacement doesn't just reset when the truck changes direction. The displacement function here is a piecewise function, one representing t = 0 to 80 and one t = 80 to 150. (though I just redefined the x-axis by saying t = 80 is t = 0 for the second piecewise function for simplicity)

displacement is the shortest distance from final position to the initial position. Therefore, while you're moving to the east, displacement is the hypotenuse of the triangle made by the north movement already made and the currently increasing east movement. This is because the hypotenuse would be the distance from the origin(initial position) to the current coordinate of the truck.

I'm uploading a picture hopefully to show you what I'm talking about:
http://img205.imageshack.us/img205/6278/68196086.jpg

As you can see, displacement becomes more than just a function of the eastward displacement. It instead becomes the hypotenuse like I've described before. And as you should know, the hypotenuse doesn't grow 1:1 with the sides of the triangle making it, so you see a sharp decline in slope despite the decently similar instantaneous velocities.

 

[ƒ(x)]

Ah ok, gotcha. Thanks for helping. Now i get to go and explain to my teacher why its 3.3