How to estimate the percentage error in speed?

Question

I travel a distance of 800meters +/- 4m in a time of 3 minutes 12 +/- 2 seconds.

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.

So far I have worked out the smallest distance of 798m and the largest distance as 802m. (4x0.05) I am not even sure if I have done that correctly?

Please help me understand what I need to do to solve this problem

Answer 1

Longest distance is 800m + 4m = 804m = nominal +0.5%

Shortest distance is 800m - 4m = 796m = nominal -0.5%

Shortest time 192s - 2s = 190s = nominal - 1.05%

Longest time 192s + 2s = 194s = nominal + 1.05%

Nominal speed = 800m / 192s = 4.1667 m/s

Fastest speed = 804m / 190s = 4.2316 m/s = nominal + 1.55%

Slowest speed = 796m / 194s = 4.1031 m/s = nominal - 1.55%

Answer 2

velocity V = L / t, or ln V = ln (L/t) = ln L - ln t

differentiate

dV/V = dL/L - dt/t

you may use this to estimate the relative error of V from those of L and t. Since distance L and time t are measured independently, the errors for L and t individually will add to the total error, therefore

DV/V = DL/L + Dt/t =4/800 + 2/192 = 0.005 + 0.0104 = 0.0154 = 1.54%

Added note

Obviously you can use Ernst S' approach to get the answer. I have to admit his solution is more to your point. What I have shown you above is what you actually use when you are dealing with real measurements in physics, or chemistry, or whatever, when you know the relative error of the control parameters and you try to work out the error on your final results (the so-called error propagation). You will come back to this when you do physics lab work.