Answer to Question #103806 in Mechanics | Relativity for Dinah

Question

The position of a particle as a function of time is given by x=(2.0m/s)t+(−3.0m/s^3)t^3. Plot x versus t for time from t=0 to t=1.0s. Find the average velocity of the particle from t = 0.35 s to t = 0.45 s . Find the average velocity of the particle from t = 0.39 s to t = 0.41 s .

Answer 1

$x(t)=2t-3t^3$

The average velocity of the particle from t = 0.35 s to t = 0.45 s :

$v_{av}=\frac{x_f-x_i}{t_f-t_i}=\frac{x(0.45)-x(0.35)}{0.45-0.35}= \frac{0.626625-0.571375}{0.10}=0.5525 \ m/s$

The average velocity of the particle from t = 0.39 s to t = 0.41 s :

$v_{av}=\frac{x_f-x_i}{t_f-t_i}=\frac{x(0.41)-x(0.39)}{0.41-0.39}= \frac{0.613237-0.602043}{0.02}=0.5597 \ m/s$