The position of a particle as a function of time is given by x = (2.0 m/s)t + (-3.0 m/s2)t2. (a) plot x-versus-t for time from t = 0 to t = 1.0 s. (do this on paper. your instructor may ask you to turn in this plot.) this answer has not been graded yet. (b) find the average velocity of the particle from t = 0.45 s to t = 0.55 s. m/s (c) find the average velocity from t = 0.49 s to t = 0.51 s.

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aristocles aristocles · Answer 1 · 2018-09-20T03:18:13+02:00

Part a)

Equation of position with time is given as

[tex]x = (2.0 m/s)t + (-3.0 m/s2)t^2[/tex]

since this equation is a quadratic equation

so it will be a parabolic graph between t = 0 to t = 1

part b)

at t = 0.45 s

[tex]x = 2* 0.45 - 3 * 0.45^2[/tex]

[tex]x_1 = 0.2925 m[/tex]

at t = 0.55 s

[tex]x = 2* 0.55 - 3*0.55^2[/tex]

[tex]x_2 = 0.1925[/tex]

now the displacement is given as

[tex]d = x_2 - x_1[/tex]

[tex]d = 0.1925 - 0.2925 = -0.1 m[/tex]

so the average velocity is given by

[tex]v = \frac{d}{t}[/tex]

[tex]v = \frac{-0.1}{0.1) = -1 m/s[/tex]

part c)

at t = 0.49 s

[tex]x = 2* 0.49 - 3 * 0.49^2[/tex]

[tex]x_1 = 0.2597 m[/tex]

at t = 0.51 s

[tex]x = 2* 0.51 - 3*0.51^2[/tex]

[tex]x_2 = 0.2397 m[/tex]

now the displacement is given as

[tex]d = x_2 - x_1[/tex]

[tex]d = 0.2397 - 0.2597 = -0.02 m[/tex]

so the average velocity is given by

[tex]v = \frac{d}{t}[/tex]

[tex]v = \frac{-0.02}{0.02) = -1 m/s[/tex]