Question

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

Answer 1

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 $v_{1}=0.21\pm 0.05\ m/s$

Final velocity $v_{2}=0.85\pm 0.05\ m/s$

Time $t = 8.0\pm 0.1\ sec$

(a). We need to calculate the average acceleration

Using formula of acceleration

$a=\dfrac{v_{2}-v_{1}}{t}$

Put the value into the formula

$a=\dfrac{0.85-0.21}{8.0}$

$a=0.08\ m/s^2$

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

Using formula of the uncertainty

$\dfrac{\Delta v}{v}=\dfrac{\Delta v}{v_{2}-v_{1}}$

Put the value into formula

$\dfrac{\Delta v}{v}=\dfrac{0.05+0.05}{0.85-0.21}$

$\dfrac{\Delta v}{v}=0.15625$

We need to calculate the uncertainty in the time

Using formula for time

$\dfrac{\Delta t}{t}=\dfrac{0.1}{8.0}$

$\dfrac{\Delta t}{t}=0.0125$

We need to calculate the uncertainty in the acceleration

Using formula for acceleration

$\dfrac{\Delta a}{a}=\dfrac{\Delta v}{v}+]\dfrac{\Delta t}{t}$

Put the value into the formula

$\dfrac{\Delta a}{a}=0.15625+0.0125$

$\dfrac{\Delta a}{a}=0.16875$

$\Delta a=0.16875\times0.08$

$\Delta a=\pm0.0135$

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

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