Question

DO A QUADRATIC FIT ON THE POSITION VS, TIME GRAPH, THE VALUE OF THE QUADRATIC "A QUAL TO HALF OF THE ACCELERATION. WHY? D A ALUINEAR FIT ON THE VELOCITY VS. TIME GRAPH. THE VALUE OF THE SLOPE IS EQUAL TO THE ACCELERATION, WHY? COMPARE THE ACCELERATION VALUES FROM BOTH GRAPHS. DO PERCENT DIFFERENCE. POST-LAB QUESTIONS: IF THE CART DID NOT ACCELERATE, WHAT WOULD BE THE SHAPES OF THE POSITION VS. TIME AND THE VELOCITY VS. TIME GRAPHS? WHY? MAKE SKETCHES OF THE 2 GRAPHS. 0.4 Eman sies, Sanart a.

need help please

Answer 1

If a particle starts from position $x_0$ with initial velocity $u$ , and acceleration $a$ (constant).

Position of particle as a function of time is $x=x_0+ut+\frac{1}{2}at^2$

During the position time plot(quadratic fit), equation used is $x=At^2+Bt+C$

Comparing the two equations, $C=x_0,B=u,A=\frac{1}{2}a$

Hence A is equal to half of the acceleration.

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When a particle moves with initial velocity $u$ , and acceleration $a$

Velocity of a particle as a function of time is $v=u+at$

For the linear fit of velocity Vs time, the equation used is $v=mt+b$

comparing the above two equations, $m=a$, $b=u$

That is slope of best fit line $m$ is equal to acceleration ($a$ ).

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Acceleration from quadratic fit is $A=\frac{1}{2}a_1=0.0731\Rightarrow a_1=0.1462\,m/s^2$

Acceleration from best fit line is $a_2=m=0.152\,m/s^2$

Percentage difference in accelerations is $\frac{|a_1-a_2|}{a_2}*100=\frac{|0.1462-0.152|}{0.152}*100=3.8\,\%$

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If the cart did not accelerate, velocity will be constant.( acceleration is zero)

$x=x_0+ut$ , $v=u$