Answer to Question #113658 in Quantum Mechanics for Mohammed Al-fatlawi

Question

A different constant velocity cart travelled 0.980.01 m in 4.20.1s.

Calculate the speed that the car traveled, including uncertainty. Show and explain ALL your working.

Answer 1

Let us consider the velocity of the cart. If $s$ is distance and $t$ is time then the velocity is

$v = \dfrac{s}{t} = \dfrac{0.98\,m}{4.2\,s} \approx 0.233\,m/s.$

Let us calculate the uncertainty of the velocity obtained above (see http://ipl.physics.harvard.edu/wp-uploads/2013/03/PS3_Error_Propagation_sp13.pdf , page 3) . First we obtain the fractional uncertainty

$\dfrac{\delta v}{v} = \sqrt{\left(\dfrac{\delta s}{s} \right)^2 +\left(\dfrac{\delta t}{t} \right)^2 } = \sqrt{\left(\dfrac{0.01\,m}{0.98\,m} \right)^2 +\left(\dfrac{0.1\,s}{4.2\,s} \right)^2 } \approx 0.0259 \approx 2.6\%.$

Next, we calculate $\delta v: \,\, \delta v = 0.0259\cdot v = 0.0259\cdot0.23\,m/s \approx 0.006\,m/s.$

So we can write $v=0.233\pm0.006\, m/s.$ However, we know $s$ and $t$ only with two significant digits, so we should round our answer and write $v = 0.23\pm0.01\,m/s.$