Un cuerpo se mueve con MRU a una rapidez de 12 m/s y recorre 240 m; luego, disminuye su ra¬pidez uniformemente a razón de 2 m/s2 hasta que adquiere una rapidez de 2 m/s2, en ese momento acelerar uniformemente a razón de 5 m/s2 hasta adquirir una rapidez de 22 m/s. Calcula el tiempo que empleó hasta dicho momento.

29 seconds

Explanation:

First we have a constant speed of 12 m/s and the distance of 240 m, so to find the time we can use the formula:

distance = speed * time

240 = 12 * time1

time1 = 20 seconds

Then, the speed decreases at 2 m/s2 until it reaches 2 m/s. So to find this time, we use this formula:

Final speed = inicial speed + acceleration * time

2 = 12 - 2 * time2

2*time2 = 10

time2 = 5 seconds.

Then, the speed increases from 2 m/s to 22 m/s with an acceleration of 5 m/s2, so we have:

Final speed = inicial speed + acceleration * time

22 = 2 + 5 * time3

5*time3= 20

time3 = 4 seconds

The total time is:

Total time = time1 + time2 + time3 = 20 + 5 + 4 = 29 seconds

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west