Calculating Average Speed: Mobile's Trip

[rAz:DD]

Homework Statement

A mobile is moving straight as following: the first portion of the road, equal to 10% of the total distance with the speed v1=10/s , and the rest with the speed v2=30m/s.
The average speed for the entire trip is?
a)25m/s b)22.5m/s c)20m/s d)15m/s

Homework Equations

\Delta v = \frac{\Delta d}{\Delta t}

The Attempt at a Solution

I calculated the times passed for both distances, replaced them in the average speed's formula and got 25 m/s, which is a wrong answer according to the key.

 

[Mark44]

I think the key is wrong. The correct approach is as you described: calculate the time needed for each portion, add them together, and divide the total distance by the total time.

First part: dist = .1D, rate = 10 m/sec, time = .1D/(10 m/sec) = .01D
Second part: dist = .9D, rate = 30 m/sec, time = .9D/(30 m/sec) = .03D

Average speed = (total distance) / (total time) = D/(.01D + .03D) = D/.04D = 1/.04 = 25 m/sec.

 

[Mark44]

BTW, the equation you show as relevant is meaningless. dv is differential velocity and d/dt is the differentiation with respect to t operator. The first represents some quantity and the second indicates an operation to perform.

 

[rAz:DD]

Thanks for the answer
And sry for that dv/dt part, i wanted to type the greek letter delta and i didn't know whether tex codes are active or not.

 

[Mark44]

Yes, they are active. For the upper case delta (looks like a triangle), use \Delta
\Delta v = \frac{\Delta d}{\Delta t}

 

[HallsofIvy]

Homework Statement

A mobile is moving straight as following: the first portion of the road, equal to 10% of the total distance with the speed v1=10/s , and the rest with the speed v2=30m/s.
The average speed for the entire trip is?
a)25m/s b)22.5m/s c)20m/s d)15m/s

Say the road is 100 m long. Then 10% is 10 m and, at 10 m/s, that requires 1 second. The second part is 90 m long and, at 30 m/s, that requires 3 seconds, making a total of 100 m in 4 seconds. That is an average speed of 25 m/s.

Homework Equations

\Delta v = \frac{\Delta d}{\Delta t}

The Attempt at a Solution

I calculated the times passed for both distances, replaced them in the average speed's formula and got 25 m/s, which is a wrong answer according to the key.

What answer does the key give?

 

[Office_Shredder]

The problem is here

v1=10/s

It's probably some really exotic unit of distance that was left out.

In other news, I'll second (or fourth at this point) the 25m/s crowd