# Basic Kinematics question

Mach
A car traveled up a hill at constant speed of 10.0 m/s and then returned down the hill at 20.0 m/s . If the time to turn around is ignored, what was the average speed for the trip?

Now i like to struggle to figure out questions myself but i can't seem to figure this one out. I have most of the kinematics formulas in front of me too. If anyone could help to give me a hint or a clue it would be great. thanks an advance.

Mentor
Try something! What's the definition of average speed? Hint: Call the distance traveled "D".

Mach
so while going down the hill would the distance then become 2D?

Mentor
If the one-way distance is D, the round trip is 2D.

Mach
ok i got the answer but i do not think that i did it the right way. please take a look at my steps

Definition:
Average Speed = [Distance Traveled]/[Time Taken to Travel that Distance]

1. I pretend the distance traveled one way up the hill is 10m, therefore 20m round trip.

2. I then figure out the average time it takes to go up and down the hill at 10m/s and and at hill at 20m/s. (20/10+20/20)/2=1.5s

3. I then use the average time and roundtrip distance to find average speed. 20/1.5=13.33m/s

Is this the correct way of solving the problem? Is my step 2 the correct way of thinking?

Bitter
You were correct in saying that the distance would be defined as 2D.

So now we must define time. We know that 2D must be divided by time, but how do we define time? Simple, were change the formula to define it as

2D all over D/v going up + D/v going down.

Do some cancellations and then plug and chug.

Mentor
ok i got the answer but i do not think that i did it the right way. please take a look at my steps

Definition:
Average Speed = [Distance Traveled]/[Time Taken to Travel that Distance]

1. I pretend the distance traveled one way up the hill is 10m, therefore 20m round trip.
That's OK. You could also stick with calling the distance D--you'll find it doesn't matter, the distance will drop out.
2. I then figure out the average time it takes to go up and down the hill at 10m/s and and at hill at 20m/s. (20/10+20/20)/2=1.5s
Think of it this way:
How much time to go up? t1 = 10/10 = 1 s
How much time to go down? t2 = 10/20 = 0.5 s
Total time = t1 + t2 = 1.5 s

3. I then use the average time and roundtrip distance to find average speed. 20/1.5=13.33m/s
Right. Total distance (20) over total time (1.5) equals average speed.

Is this the correct way of solving the problem? Is my step 2 the correct way of thinking?