Let be $v(t, x) =\frac{x}{t}$ the average speed function, we calculate the uncertainty in the average speed by total differentials, which is in this case:

$\Delta v = \frac{\partial v}{\partial x}\cdot \Delta x+\frac{\partial v}{\partial t}\cdot \Delta t$

Where:

$\Delta v$ - Uncertainty in the average speed, measured in meters per second.

$\frac{\partial v}{\partial x}$ - Partial derivative of the average speed function with respect to distance, measured in $s^{-1}$ .

$\frac{\partial v}{\partial t}$ - Partial derivative of the average speed function with respect to time, measured in meters per square second.

$\Delta x$ - Uncertainty in distance, measured in meters.

$\Delta t$ - Uncertainty in time, measured in seconds.

Partial derivatives are, respectively:

$\frac{\partial v}{\partial x} = \frac{1}{t}$ , $\frac{\partial v}{\partial t} = - \frac{x}{t^{2}}$

Then, the total differential expression is expanded as:

$\Delta v = \frac{\Delta x}{t}-\frac{x\cdot \Delta t}{t^{2}}$

If we get that $\Delta x = 2.3\,m$ , $t = 6.3\,s$ , $x = 6.1\,m$ and $\Delta t = 1.5\,s$ , the uncertainty in the average speed is:

$\Delta v = \frac{2.3\,m}{6.3\,m}-\frac{(6.1\,m)\cdot (1.5\,s)}{(6.3\,s)^{2}}$

$\Delta v = 0.134\,\frac{m}{s}$

The uncertainty in the average speed is 0.134 meters per second.

8 0

4 years ago

Evaluate the expression for s = 2.<br> 9s- 2(s+ 1)2 =

igor_vitrenko [27]

Answer:

s= $\frac{40}{21}$

Step-by-step explanation:

7 0

4 years ago

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