Calculating Average Speed: A Stupid Speed Question Homework Solution

[IronBrain]

Homework Statement

A car travels up the hill at a constant speed of 30 km/h, then returns down hill at the speed 66 km/h. Calculate the average speed round trip

I can't believe I am stumped on such a kindergarten problem, I hate online homework and the garbage system they are designed on.

Homework Equations

The Attempt at a Solution

\frac{66+30}{2} =48 km/h

Says I am wrong, which I prolly am, then states the average speed is the ratio of the total displacement over time, obviously, then says use to D to symbolize distance if need be...?

 

[berkeman]

Homework Statement

A car travels up the hill at a constant speed of 30 km/h, then returns down hill at the speed 66 km/h. Calculate the average speed round trip

I can't believe I am stumped on such a kindergarten problem, I hate online homework and the garbage system they are designed on.

Homework Equations

The Attempt at a Solution

\frac{66+30}{2} =48 km/h

Says I am wrong, which I prolly am, then states the average speed is the ratio of the total displacement over time, obviously, then says use to D to symbolize distance if need be...?

I think you just need to pick an arbitrary distance and calculate how long each leg of the trip takes, and use those numbers to figure out the average speed...

 

[rogerbacon]

Another hint: The way you average is as if the car was going the same amount of time at 30km/h and 66km/h.

 

[berkeman]

Another hint: The way you average is as if the car was going the same amount of time at 30km/h and 66km/h.

Really? But it isn't taking the same amount of time of course, since it it going the same distance at the two speeds in this question. Still, I'll wait until the OP posts his work to see if your way works too.

 

[IronBrain]

Hmm as if the car is going the same amount of time at both speeds...
This question is just a tad confusing, pick an arbitrary distance to find the how long each trip takes

something like this? Say I pick 25 km for the distance

30\frac{km}{h} = \frac{25 km}{t}

 

[berkeman]

This question is just a tad confusing, pick an arbitrary distance to find the how long each trip takes

something like this? Say I pick 25 km for the distance

30\frac{km}{h} = \frac{25 km}{t}

Yeah. Find t1 and t2 to get your total time, and you know your total distance... And then check it again for some other arbitrary distance. Do you get the same average speed?

 

[IronBrain]

Ok, let's see here for the first constant speed of 30 km/h, I am going to pick an arbritary distance of 25 km for both speed constants

Speed:30 km/h

Distance: 25

Time: 0.83

Next speed constant

Speed: 66 km/h

Distance: 25

Time:0.3787

Average Speed
\frac{25}{1.2087}=20.68 km/h

Picking another arbitrary distance say 60 km

My average speed equates to 20.62 km/h

Pretty close I'd say, and all I had to do is that? Wow, and the keyword in the problem was "return"?

 

[berkeman]

Good job!