% 1 a Bereken de calibratiecurve
m = 0:5:20;
g = 9.8;
F = m*g;
y = [12 777 1590 2353 3128];
coef = polyfit (F,y,1);
% 1 b Bereken welke waarde uit y = 2012 komt
x = 1/coef(1)*(2012-coef(2));
% 1 c Bereken de kalibratiecurve waar geen gewicht en 20 kg werd
geleverd
m = [0 20];
g = 9.8;
F = m*g;
y = [12 3128];
coef1 = polyfit (F,y,1);
% 1 d Bereken uit kalibratiecurve c wat uit waarde y =2012 komt
x1 = 1/coef1(1) * (2012-coef1(2));
% 1 e welke waarde zal waarschijnlijk dichter bij de waarheid zitten
b of d
b zal waarschijnlijk dichter bij zitten omdat hier gebruik is
gemaakt van meerdere gewichten (meer dan 2)
% 2 a Is het goed gekalibreerd en wat kan beter
De kalibratie is goed opgesteld, alleen hadden meer meetpunten dan 2
gebruikt kunnen worden
% 2 b Bepaal de kalibratiecurve van de lengte en breedte van de
tafel
lengte = 284;
breedte = 142;
hoek1 = [122;421];
hoek2 = [2688;421];
hoek3 = [122;1703];
oorsprong = hoek1;
pixbreedte = (hoek2 - hoek1)/lengte;
pixlengte = (hoek3 - hoek1)/breedte;
% 3 e Bereken de 95% CI
Ondergrens = Gemiddelde -1.96*SEm;
Bovengrens = Gemiddelde +1.96*SEm;
CI = [Ondergrens Bovengrens];
% 4 a bepaal de CI van de x coordinaat op 99%
x = [41.283 41.034 41.064 40.958 41.022 40.914];
Gem = mean (x);
SD = std (x);
SEm = SD/(sqrt(length(x)));
Ondergrens = Gem - 2.58*SEm;
Bovengrens = Gem + 2.58*SEm;
CI = [Ondergrens Bovengrens];
% 4 b Is de relatieve error hier belangrijk, waarom?
Nee de relatieve error is hier niet belangrijk omdat er niet
gescoord is op een ratio schaal (geen absolute 0)
% % 5 Geef de snelheid in m/s met de absolute error
V_kmh = 59;
M_s = 59/3.6;
rel_error = 5;
abs_error = M_s/100*5;
% 6 Bereken de versnelling van het object en de absolute en
relatieve error
F = 154;
m = 80.3;
a = F/m;
abs_error_F = 2;
abs_error_m = 0.1;
abs_error_a = sqrt (((1/m)^2 * (abs_error_F)^2) + (((-F*m^-
2)^2*(abs_error_m)^2)));
rel_error_a = (abs_error_a/a)*100;
LET OP HIERBIJ MOET JE PARTIEEL DIFF:
1 1
A = F/m -> F* -> over F diff =
m m
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