Suppose the position vector for a particle is given as a function of time...

Question

by r(t) = x(t)î + y(t)ĵ, with x(t) = at + b and y(t) = ct² + d, where a = 2.00 m/s, b = 1.30 m, c = 0.132 m/s², and d = 1.20 m.

(a) Calculate the average velocity during the time interval from t = 2.05 s to t = 3.95 s.

(b) Determine the velocity at t = 2.05 s.

(c) Determine the speed at t = 2.05 s.

Answer 1

It has been many years since I studied this in graduate school, so I can't guarantee the solutions are correct. Check out my solutions to see if they make sense.

(a) The average velocity between the two times is the vector difference of the positions vectors at the two times divided by the difference in time. Substitute the x(t) and y(t) functions in the position vector equation, enter the times to get the position vectors at each time. Subtract the position vectors for the two times and divide by the difference in time. I got V_avg = 2.00i + 0.79j

(b) The velocity at 2.05s was obtained by differentiating the position vector equation with respect to time after putting in the x(t) and y(t) functions. I obtained V_t=2.05 = 2.00i + 0.54j.



(c) The speed is just the magnitude of the velocity vector at 2.05 s. This is the square root of the sum of the squares of the coefficients of i and j in the velocity expression. I got 2.07 m/s.

Please check this out to see if it makes sense. Also, I presume you are taking an advanced level course in physics, and I would like to know what the actual solutions are. Thanks.