Why Does Average Speed Calculation Differ for Uphill and Downhill Travel?

[guppster]

Homework Statement

A Bus drives 1km up a hill in 5.0 minutes. It then drives down the hill in 4.0 minutes. For the bus, find
a) the average speed up the hill
b) the average speed down the hill
c)the average speed for the whole trip
d)why is the answer for c) not equal to (speed up the hill + speed down the hill) / 2

Homework Equations

v = d / t

The Attempt at a Solution

a) 1000/300 = 3.33 m/s
b) 1000/240 = 4.16 m/s
c) (1000 + 1000) / (300 + 240) = 3.70 m/s
d) *This is what i do not understand* The formula i am suppose to prove incorrect returns the same result as my calculation. Are my previous calculations incorrect, or are they suppose to be equal.
ie. (3.33 + 4.16)/2 = 3.745 m/s

Thanks,

 

[cepheid]

Welcome to PF,

Homework Statement

A Bus drives 1km up a hill in 5.0 minutes. It then drives down the hill in 4.0 minutes. For the bus, find
a) the average speed up the hill
b) the average speed down the hill
c)the average speed for the whole trip
d)why is the answer for c) not equal to (speed up the hill + speed down the hill) / 2

Homework Equations

v = d / t

The Attempt at a Solution

a) 1000/300 = 3.33 m/s
b) 1000/240 = 4.16 m/s
c) (1000 + 1000) / (300 + 240) = 3.70 m/s
d) *This is what i do not understand* The formula i am suppose to prove incorrect returns the same result as my calculation. Are my previous calculations incorrect, or are they suppose to be equal.
ie. (3.33 + 4.16)/2 = 3.745 m/s

Thanks,

They're NOT equal. 3.70 m/s ≠ 3.75 m/s

 

[guppster]

Welcome to PF,



They're NOT equal. 3.70 m/s ≠ 3.75 m/s

Thank you for welcoming me to PF.
I am truly very grateful you replied.

Is there a valid reason for this inequality, i think it is because the velocity values are not very precise.

 

[cepheid]

Thank you for welcoming me to PF.
I am truly very grateful you replied.

Is there a valid reason for this inequality, i think it is because the velocity values are not very precise.

Nope, that's not the reason. These two averages are not supposed to be the same, unless the car spends equal time covering both distances, which it doesn't. You can show this mathematically. If v₁ = d₁/t₁ and v₂ = d₂/t₂, and v = (d₁ + d₂)/(t₁ + t₂), how does v compare to (1/2)(v₁ + v₂) algebraically? Are they the same?

 

[guppster]

Nope, that's not the reason. These two averages are not supposed to be the same, unless the car spends equal time covering both distances, which it doesn't. You can show this mathematically. If v₁ = d₁/t₁ and v₂ = d₂/t₂, and v = (d₁ + d₂)/(t₁ + t₂), how does v compare to (1/2)(v₁ + v₂) algebraically? Are they the same?

Um, No they are not the same.

 

[cepheid]

Um, No they are not the same.

Did you work it out? Can you show me your work? Plug in the expressions for v in terms of d and t for each of the velocities (v1, v2).

 

[guppster]

v1 = d1/t1
v2 = d2/t2

like this?

(1/2)(d1/t1 + d2/t2)

 

[cepheid]

v1 = d1/t1
v2 = d2/t2

like this?

(1/2)(d1/t1 + d2/t2)

Yeah exactly. Now, if you compare that to v = (d1 + d2)/(t1 + t2), you can easily see that these two expressions are not the same, *unless* if t1 = t2, in which case they become the same.