Answer:
The uncertainty in the average speed is 0.134 meters per second.
Step-by-step explanation:
Let be [tex]v(t, x) =\frac{x}{t}[/tex] the average speed function, we calculate the uncertainty in the average speed by total differentials, which is in this case:
[tex]\Delta v = \frac{\partial v}{\partial x}\cdot \Delta x+\frac{\partial v}{\partial t}\cdot \Delta t[/tex]
Where:
[tex]\Delta v[/tex] - Uncertainty in the average speed, measured in meters per second.
[tex]\frac{\partial v}{\partial x}[/tex] - Partial derivative of the average speed function with respect to distance, measured in [tex]s^{-1}[/tex].
[tex]\frac{\partial v}{\partial t}[/tex] - Partial derivative of the average speed function with respect to time, measured in meters per square second.
[tex]\Delta x[/tex] - Uncertainty in distance, measured in meters.
[tex]\Delta t[/tex] - Uncertainty in time, measured in seconds.
Partial derivatives are, respectively:
[tex]\frac{\partial v}{\partial x} = \frac{1}{t}[/tex], [tex]\frac{\partial v}{\partial t} = - \frac{x}{t^{2}}[/tex]
Then, the total differential expression is expanded as:
[tex]\Delta v = \frac{\Delta x}{t}-\frac{x\cdot \Delta t}{t^{2}}[/tex]
If we get that [tex]\Delta x = 2.3\,m[/tex], [tex]t = 6.3\,s[/tex], [tex]x = 6.1\,m[/tex] and [tex]\Delta t = 1.5\,s[/tex], the uncertainty in the average speed is:
[tex]\Delta v = \frac{2.3\,m}{6.3\,m}-\frac{(6.1\,m)\cdot (1.5\,s)}{(6.3\,s)^{2}}[/tex]
[tex]\Delta v = 0.134\,\frac{m}{s}[/tex]
The uncertainty in the average speed is 0.134 meters per second.
