# How to calculate the percentage error in speed

1. A person travels 800m +/-4m in a time of 3 minutes 12 seconds +/-2sec

I have to work out the largest and smallest possible values for the speed along with the percentage error for the speed and distance travelled.

2. Homework Equations = Distance/Time = Speed

3. The Attempt at a Solution

This is were I get myself in a muddle I am not sure if I should add +/-4m to 800m to make it 804m and 796m or should it be 798m and 802m = +/-4m (like wise with the time) If someone could explain the correct steps in solving this problem that would be great.

## Answers and Replies

LowlyPion
Homework Helper
1. A person travels 800m +/-4m in a time of 3 minutes 12 seconds +/-2sec

I have to work out the largest and smallest possible values for the speed along with the percentage error for the speed and distance travelled.

2. Homework Equations = Distance/Time = Speed

3. The Attempt at a Solution

This is were I get myself in a muddle I am not sure if I should add +/-4m to 800m to make it 804m and 796m or should it be 798m and 802m = +/-4m (like wise with the time) If someone could explain the correct steps in solving this problem that would be great.

+ / - means just that. 804 and 796 are the upper and lower bound on possible distances measured. What is the spread then in "speeds" using 190 and 194 seconds as the time traveled?

+ / - means just that. 804 and 796 are the upper and lower bound on possible distances measured. What is the spread then in "speeds" using 190 and 194 seconds as the time traveled?

I calculated these figures...

Largest average speed 804/190 = 4.23m/s

Smallest average speed 796/194 = 4.10m/s

Now what formula should I use to calculate the percentage errors?

LowlyPion
Homework Helper
I calculated these figures...

Largest average speed 804/190 = 4.23m/s

Smallest average speed 796/194 = 4.10m/s

Now what formula should I use to calculate the percentage errors?

I'd split the difference between high and low and then express that difference as a percentage + / - off the average of the two high and low values.

What about using the nominal values given and the speed based on the nominal values as the denominator?

I was just doing some further reading on the formula would that be; ERROR/DISTANCE or TIME x 100 ??

Longest Distance = +4m divided by 800m multiply by 100 = +0.5%

Longest Time = +2s divided by 192s multiply by 100 = +1.041% or +1.05%

Have I got my calculations correct ? Rounding to 1 significant figure would be 1.05% right?

First off, I will say I don't know for sure what the answer should be. I was taught that percentage error was abs( assumed answer - true value) / true value *100%.

I would agree with your distance percentage. The assumed answers would be 800+/-4, or 796 to 804. I would think the true answer would be 800. So using either 804 or 796 would give either abs(4)/800 or abs(-4)/800, x 100%.

You have calculated percentage error for time. Your problem says to find for speed. I would agree with your answer though, except I wouldn't round the figures as you did. Do you think with a "1" after the "4" it should be rounded up to "5", or kept as "4"?

The trickier part is speed. Use your equation for speed. You have a smallest number and biggest number for distance and a smallest and biggest number for time. If you do all possible arrangements you will calculate four speeds. Take the smallest and biggest as the range of speeds. If you use some properties of fractions you could do just two calculations to get the smallest and biggest without doing all four.

Now that you have the smallest and biggest, calculate the true speed using the 800 and 192. Use the range of speeds and this true speed to find a percentage error. The problem is that you will get a different answer if you use the upper speed to calculate percentage error than if you use the lower speed. I would probably put down the higher calculated error as my answer.

However, my advise to you would be to show all work. Show that you did the calculation for both higher and lower speed and picked the higher one as the answer. If the teacher doesn't agree, you will hopefully get most of the credit for having calculated both.

The other possible method would be that you are expected to take the biggest value - smallest value and then divide by the true answer. But this doesn't seem to exactly fit the definition of percentage error that I learned. Maybe one of the homework helpers here can give an indication of which way it should be.

Last edited:
LowlyPion
Homework Helper
What about using the nominal values given and the speed based on the nominal values as the denominator?

Yes, I believe you are right. That is a better way to use the nominal velocity determined by the nominal values and the quotient rule to determine the resulting relative error which is probably the answer being sought by the problem.

The quotient rule for error expression/propagation I think is the square root of the sum of the squares of the individual nominal errors.

In this case I think it would be the sqrt( 4/800 + 2/192 ) expressed as a percentage of the nominal velocity 800/192.

I imagine that since the error in velocity is also less than just the simple sum of the errors i.e. 4/800 + 2/192 that could also be used.

Whatever method explain your work to the grader.