How to find final speed when initial speed and average speed is given

Unknown008 · Jul 27, 2010, 05:54 AM

Assuming the acceleration is constant;

Remember that:

$Average\ Speed = \frac{Total\ Distance}{Total\ time}$

Or

$V = \frac{s}{t}$

And use: $v^2 = u^2 + 2as$

And

$v = u+at$

Post what you come up with

Dinsdale1963 · Jul 28, 2010, 06:55 PM

you'll have a difficult time using the previous equation, since you're not given time or acceleration. A much simpler approach is to us the formula:

avg. V = (v+u)/2
Therefore,

u = 2(avg v) - v
So...

u = (2)(48) 40 = 58 m/s

Unknown008 · Jul 29, 2010, 01:41 AM

I usually don't use this formula, that's why I posted the 'long way round'. But at least, I'm sure what I'm doing. If you have got a shorter, easier alternate answer, that's fine too

ebaines · Jul 29, 2010, 06:01 AM

Quote:

Originally Posted by Dinsdale1963

avg. V = (v+u)/2

Sorry - but this is not right! The average speed to cover the distance out and back is not equal to the arithmetic average of the two speeds, but rather is equal to the total distance traveled divided by the total time it takes. Let d = the distance out, so that the total distance traveled is 2d, and the times for the drive out and the drive back back are $t_1$ and $t_2$ respectively:

$ V_{avg}= \frac {Total \ Distance} {Total \ Time} = \frac {2d} {t_1 + t_2} $

The times for each leg of the trip are
$ T_1 = \frac d {v} \\ T_2 = \frac d {u} $ .

Thus:

$ T_1 + t_2 = \frac d {v} + \frac d {u} $

Put this back into the first equation:

$ V_{avg}= \frac {2d} {\frac d {v} + \frac d {u}} \\ V_{avg}= \frac {2 v u} {v + u} $

Rearrange and you get

$ U = \frac {v_{avg}v} {2v-v_{avg}} = \frac {48 \times 40} {2 \times 40 - 48} = 60 m/s $

To check that this is correct, try using a value for d that makes the math easy: let's assue d = 12000 m. The total time to make the trip out and back is 12000m/(40m/s) + 12000m/(60m/s) = 500 seconds. The total distance covered is 24000m, so the average speed is 24000m/500s = 48 m/s. So it checks.

ashpar_kantan · Jul 30, 2010, 02:14 AM

Quote:

Originally Posted by ebaines

Sorry - but this is not right! The average speed to cover the distance out and back is not equal to the arithmetic average of the two speeds, but rather is equal to the total distance traveled divided by the total time it takes. Let d = the distance out, so that the total distance traveled is 2d, and the times for the drive out and the drive back back are $t_1$ and $t_2$ respectively:

$ V_{avg}= \frac {Total \ Distance} {Total \ Time} = \frac {2d} {t_1 + t_2} $

The times for each leg of the trip are
$ T_1 = \frac d {v} \\ T_2 = \frac d {u} $ .

Thus:

$ T_1 + t_2 = \frac d {v} + \frac d {u} $

Put this back into the first equation:

$ V_{avg}= \frac {2d} {\frac d {v} + \frac d {u}} \\ V_{avg}= \frac {2 v u} {v + u} $

Rearrange and you get

$ U = \frac {v_{avg}v} {2v-v_{avg}} = \frac {48 \times 40} {2 \times 40 - 48} = 60 m/s $

To check that this is correct, try using a value for d that makes the math easy: let's assue d = 12000 m. The total time to make the trip out and back is 12000m/(40m/s) + 12000m/(60m/s) = 500 seconds. The total distance covered is 24000m, so the average speed is 24000m/500s = 48 m/s. So it checks.

Thanks. That was really helpful.

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