La ecuación de la posición de una esferita está dada por: r(t)=(2.Cos(πt) i-3.Sen(πt) j) (m) ¿Cuál es la velocidad de la esferita a los 0,25 segundos? Calcule la aceleración media en los primeros tres segundos

v = (-4.44 i^ + 6.66 j^ )  m/s, a_average =( 0 i^ -2π j^) m/s²

Explanation:

The expression left corresponds to an oscillatory movement (MAS), the speed is defined by

         v = dr / dt

the function of position

         r = 2 cos πt  i^  + 3 sin πt  j^

let us note that it is a movement in two dimensions

let's perform the derivative

          v = -2π sin πt  i^  + 3π cos πt  j^

we evaluate this expression for t = 0.25 s, remember that the angle is in radians

          v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^

           v = (-4.44 i^ + 6.66 j^ )  m/s

To calculate the mean acceleration we use the expression

           a = () / Δt

indicates that the time is the first 3 s

we look for the initial velocity t = 0 s

           v₀ = 0 i ^ + 3π j ^

we look for the fine velocity, t = 3 s

          v_f = - 2π sin (π 3) + 3π cos (π 3) j ^

          v_f = 0 i ^ - 3π j ^

we calculate the average acceleration

            Δt = (3 -0) = 3 s

           a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3

           a_average = (0 i ^ -2π j ^ ) m/s²

this could be newton#'s law of cooling. the cooling curve is exponential

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)