Question

Value Position vs Time Linear Fit m y = mx + b Quadratic Fit Α. B 0.305 -0.0583 y=0.305x -0.0583 4 1 0.110 -0.0663 0.195... yFor Position vs Time data: (a) Did your quadratic fit of this graph provide initial position? If yes, what is its value? (4 pPlease note that X is time

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Answer #1

1.

Yes it is.

given,

r = 0.11t- 0.0663t+0.195

initially at t=0 sec

r = 0.11 * 0 - 0.0663* 0 + 0.195

x=0.195m

2.

Yes it is.

velocity:
v=\frac{\mathrm{d} x}{\mathrm{d} t}

(0.11 -0.063 +0.195)

v= 2*0.11*t -0.0663

initially at t= 0 sec

V = -0.0663m/s

3.

yes it is

acceleration:

du

A= (0.22 -0.0663

a = 0.22m/s

4.

slope

m= \frac{x_2-x_1}{t_2-t_1}

m = meter/sec

its a unit of velocity.

also tܠ|ܠ will give velocity

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