value 2. For Position vs Time data: (a) Did your quadratic fit of this graph pro

Question

value

2. For Position vs Time data:

(a) Did your quadratic fit of this graph provide initial position? If yes, what is its value? (4 points)

(b) Did your quadratic fit of this graph provide initial velocity? If yes, what is its value? (4 points)

(c) Did your quadratic fit of this graph provide acceleration? If yes, what is its value?

(4 points) (d) What specific physical quantity does the slope of the two middle points from the Position vs. Time graph represent?

3. For Velocity vs Time data:

(a) Did your linear fit of this graph provide initial position? If yes, what is its value? (4 points)

(b) Did your linear fit of this graph provide initial velocity? If yes, what is its value? (4 points)

(c) Did your linear fit of this graph provide acceleration? If yes, what is its value? (4 points)

(d) How does the time your calculated average velocity occurred at compare to the times of the two middle points from the position vs. time graph? (3 points)

(e) How does the time your calculated average velocity value occurred at relate to the time values of the first and last good data points in the Velocity vs. Time graph? (3 points)

Postition vs Time

value

linear fit m 0.168 B -0.474 y=mx+b 0.168t+(-0.474) Quadratic fit A 0.452 B -0.317 C 0.743 y=Ax^2+Bx+C y=(0.432)t^2+(-0.317)t+(0.743) (X1,Y1) 3.200,0.182 (X2,Y2) 6.200,0.512 slope 0.32m/s Velocity VS Time Value m 0.0872 b -0.297 y=mx+b y=(0.0872)t+(-0.297) (X1,Y1) 3.500,0.01 (X2,Y2) 7.500,0.37 Vavg Vavg time Acceleration VS Time Value m 0.00756 b 0.0428 y=mx+b y=(0.00756)t+(0.0428) The Mean 0.084

Explanation / Answer

a) Yes we can get the initial position by the constant value i.e. C.

y0 = 0.743m

b) yes

the coefficient of t gives value of initial velocity i.e B

v0 = -0.317m/s

c) yes

it is equal to half of coefficient of t^2 i.e. A/2 = 0.432/2 = 0.216m/s^2

d) This quantity represent the average velocity

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