(4-1) Please show all work! A particle moves in one dimension, and its position

Question

(4-1) Please show all work!

A particle moves in one dimension, and its position as a function of time is given by x (1.8 m/s)t (-3.4 m/s2)t2 (a) What is the particle's average velocity from t 0.45 s to t 0.55 s? (Indicate the direction with the sign of your answer. 1.46 Average velocity is the change in position over change in time. Think about how to use the function to find the change in position. m/s (b) What is the particle's average velocity from t 3 0.49 s to t 0.51 s? (Indicate the direction with the sign of your answer.) m/s

Explanation / Answer

X = 1.8t -3.4 t^2

x(0.45) = 1.8*0.45-3.4*0.45^2 = 0.1215m

x(0.55) = 1.8*0.55-3.4*0.55^2 = -0.0385m

Va = (x(0.55)-x(0.45)/(0.55-0.45) = -0.16/0.1 = -1.6 m/s

x(0.49) = 1.8*0.49-3.4*0.49^2 = 0.06566 m

x(0.51) = 1.8*0.51-3.4*0.51^2 = 0.03366

Va = (x0.51)-x(0.49))/(0.51-0.49) = (0.03366-0.06566)/0.02 = -1.6 m/s

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