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To find the acceleration of a glider moving down a sloping air track, you measure its velocities (V1 and V2) at two points and the time t it takes between them, as follows:

V1=0.21+/- 0.05m/s, V2=0.85 +/- 0.05m/s, t=8.0 +/- 0.1s
a) Assuming all uncertainties are independent and random, find the average acceleration, a=(v2-v1)/t , and its uncertainty.
b) How well does your result in part (a) agree with your friends theoretical prediction that a=0.13 +/- 0.01m/s^2

Respuesta :

Answer:

(a). The average acceleration is 0.08 m/s².

(b). The uncertainty in the acceleration is ±0.0135 m/s².

Explanation:

Given that,

Initial velocity [tex]v_{1}=0.21\pm 0.05\ m/s[/tex]

Final velocity [tex]v_{2}=0.85\pm 0.05\ m/s[/tex]

Time [tex]t = 8.0\pm 0.1\ sec[/tex]

(a). We need to calculate the average acceleration

Using formula of acceleration

[tex]a=\dfrac{v_{2}-v_{1}}{t}[/tex]

Put the value into the formula

[tex]a=\dfrac{0.85-0.21}{8.0}[/tex]

[tex]a=0.08\ m/s^2[/tex]

(b). We need to calculate the uncertainty in the velocity

Using formula of the uncertainty

[tex]\dfrac{\Delta v}{v}=\dfrac{\Delta v}{v_{2}-v_{1}}[/tex]

Put the value into formula

[tex]\dfrac{\Delta v}{v}=\dfrac{0.05+0.05}{0.85-0.21}[/tex]

[tex]\dfrac{\Delta v}{v}=0.15625[/tex]

We need to calculate the uncertainty in the time

Using formula for time

[tex]\dfrac{\Delta t}{t}=\dfrac{0.1}{8.0}[/tex]

[tex]\dfrac{\Delta t}{t}=0.0125[/tex]

We need to calculate the uncertainty in the acceleration

Using formula for acceleration

[tex]\dfrac{\Delta a}{a}=\dfrac{\Delta v}{v}+]\dfrac{\Delta t}{t}[/tex]

Put the value into the formula

[tex]\dfrac{\Delta a}{a}=0.15625+0.0125[/tex]

[tex]\dfrac{\Delta a}{a}=0.16875[/tex]

[tex]\Delta a=0.16875\times0.08[/tex]

[tex]\Delta a=\pm0.0135[/tex]

Hence, (a). The average acceleration is 0.08 m/s².

(b). The uncertainty in the acceleration is ±0.0135 m/s².

pristina

(a) The average acceleration obtained is [tex]a= (0.08 \pm 0.02)m/s^2[/tex]

(b) The acceleration predicted in part-b is 1.6 times more than that obtained in part-a. Also, the uncertainty in acceleration predicted in part-b is half of that obtained in part-a.

Uncertainty in Measurements

Given, that the initial and final velocities are;

[tex]v_1 = (0.21 \pm0.05)\,m/s\\v_2 = (0.85 \pm0.05)\,m/s[/tex]

Also, the time is given as;

[tex]t = (8.0 \pm0.1)\,s[/tex]

Now the average acceleration is given by;

[tex]a = \frac{v_2 - v_1}{t} =\frac{0.85\,m/s-0.21\,m/s}{8\,s} =0.08\,m/s^2[/tex]

Now calculating the percentage of uncertainties in each variable, we get;

[tex]\%u_{v_1} = \frac{\Delta v_1}{v_1} \times 100 = 23.8\%\\\%u_{v_2} = \frac{\Delta v_2}{v_2} \times 100 = 5.88\%\\\%u_{t} = \frac{\Delta t}{t} \times 100 = 1.25\%[/tex]

Therefore, the percentage of the uncertainty in acceleration is obtained by adding the percentage uncertainties in initial and final velocities and time.

[tex]\% u_{a} = 5.88 \% +23.8 \% + 1.25\% = 30.93 \%[/tex]

Now the uncertainty in acceleration is calculated by;

[tex]\Delta a = a \times \frac{\% u_{a}}{100} =0.08 \times\frac{30.93}{100} = 0.02[/tex]

Therefore the acceleration can be written as;

[tex]a= (0.08 \pm 0.02)m/s^2[/tex]

(b) The acceleration predicted in part-b is 1.6 times more than that obtained in part-a. Also, the uncertainty in part-b is half of that obtained in part-a.

Learn more about uncertainty here:

https://brainly.com/question/13318624

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