Calculating the Average speed given two speeds

[jnuz73hbn]

Homework Statement

Hello, I have 2 speeds and need to calculate the average speed during the time traveled

Relevant Equations

Is it possible to solve this with:

$$ v_1= 115km/h $$ $$ v_2= 60km/h $$ solution: $$ \frac {v_1 + v_2}{2} $$

 

[Hill]

No, this is not the average speed during the time travelled.

 

[BvU]

Homework Statement: Hello, I have 2 speeds and need to calculate the average speed during the time traveled

You have two speeds at the same time ?

 

[jnuz73hbn]

only one object that first moves at $$60km/h$$ and then at $$115km/h$$

 

[DaveE]

You haven't really stated the full question. Average is a mathematical operation that can be applied to different problems in different ways. So, this may make sense if you are asking about the average result from a bunch of separate cars on a race track. But for one car driving on a highway the average speed for the whole trip is calculated from the total distance divided by the total time. For example, suppose I can drive for 10 minutes at 60mph (=10 miles) but then have to drive for 50 minutes at 30mph (=25 miles). My average speed would then be 35 miles in 1 hour.

Speed is distance divided by time. Normally you have to define what each of those are before you do the speed calculation.

 

[jnuz73hbn]

You haven't really stated the full question. Average is a mathematical operation that can be applied to different problems in different ways. So, this may make sense if you are asking about the average result from a bunch of separate cars on a race track. But for one car driving on a highway the average speed for the whole trip is calculated from the total distance divided by the total time. For example suppose I can drive for 10 minutes at 60mph (=10 miles) but then have to drive for 50 minutes at 30mph (=25 miles). My average speed would then be 35 miles in 1 hour.

i know that, but I only have two 2 speeds for a car, 60kmh in the first half of the journey and 115km/h in the second half

 

[haruspex]

only one object that first moves at $$60km/h$$ and then at $$115km/h$$

Always start with the definitions. What is the definition of average speed?

 

[jnuz73hbn]

What is the definition of average speed?

$$ v= \frac{s}{t} $$ but I dont have $$s$$ I only have two 2 speeds for a car, 60kmh in the first half of the journey and 115km/h in the second half

 

[DaveE]

i know that, but I only have two 2 speeds for a car, 60kmh in the first half of the journey and 115km/h in the second half

"half of the journey" is an important piece of information. Make up a value for total length for the trip and do the calculation. Then try it again for a different choice of total length. Does the length matter?

Or maybe "half" means half of the time. Try with different total times for the trip.

 

[jnuz73hbn]

"half of the journey" is an important piece of information. Make up a value for total length for the trip and do the calculation. Then try it again for a different choice of total length. Does the length matter?

Or maybe "half" means half of the time. Try with different total times for the trip.

Nothing is given about the route, you should only calculate using this information. It means half of the time.

 

[DaveE]

Nothing is given about the route, you should only calculate using this information.

The route doesn't matter. Turns don't matter. The only variables that go into a speed calculation are distance and time.

 

[PeroK]

i know that, but I only have two 2 speeds for a car, 60kmh in the first half of the journey and 115km/h in the second half

If you drive halfway round a racetrack at a speed of 60 km/h and at a speed of 115 km/h for the the second half, then the average speed is not $\frac{60 + 115} 2$ km/h. That a common mistake.

 

[Orodruin]

$$ v= \frac{s}{t} $$ but I dont have $$s$$ I only have two 2 speeds for a car, 60kmh in the first half of the journey and 115km/h in the second half

"Half the journey" is not very specific. It could mean "half of the time" or "half the distance". Each of those will give different results for the average speed. You need to specify which one you mean.

 

[jnuz73hbn]

If you drive halfway round a racetrack at a speed of 60 km/h and at a speed of 115 km/h for the the second half, then the average speed is not $\frac{60 + 115} 2$ km/h. That a common mistake.

but look, at the half of the driving time, speed is 60km/h, in the second half driving time, speed = 115km/h.

 

[jnuz73hbn]

"Half the journey" is not very specific. It could mean "half of the time" or "half the distance". Each of those will give different results for the average speed. You need to specify which one you mean.

first half of time=60km/h; second half of time is 115km/h

 

[PeroK]

first half of time=60km/h; second half of time is 115km/h

So, if the journey takes two hours, then the first half of the journey is 60 km and the second half of the journey is 115 km?

 

[Orodruin]

first half of time=60km/h; second half of time is 115km/h

So, assume that the full time is $T$. How far do you get in the total journey? The average speed is that distance divided by $T$.

For the other case, assume that the total distance is $D$. How long will it the journey take travelling $D/2$ at the first speed and then $D/2$ at the second speed? The average speed is then $D$ divided by that time.

I suggest you perform both of these computations to find out that the result is not the same.

 

[jnuz73hbn]

$$t_1=\frac{D/2}{v_1}$$
that means:
$$v=\frac{2}{1/v_1+1/v_2}$$

 

[jnuz73hbn]

So, assume that the full time is $T$. How far do you get in the total journey? The average speed is that distance divided by $T$.

For the other case, assume that the total distance is $D$. How long will it the journey take travelling $D/2$ at the first speed and then $D/2$ at the second speed? The average speed is then $D$ divided by that time.

I suggest you perform both of these computations to find out that the result is not the same.

is this new formula correct?

 

[haruspex]

$$t_1=\frac{D/2}{v_1}$$
that means:
$$v=\frac{2}{1/v_1+1/v_2}$$

That is correct if the two distances are the same. This expression is known as the "harmonic mean".

 

[Orodruin]

is this new formula correct?

For the case of travelling half the distance at each speed, yes. Note that it can also be written:
$$
\frac 1v = \frac 12\left( \frac 1{v_1} + \frac 1{v_2}\right)
$$
ie, the reciprocal of the mean speed is the average of the reciprocals of the two speeds.

The case if half the time at each velocity is different as already mentioned.