Un ciclista en ultimo tramo de una prueba deportivas mueve con aceleración constante y cubre entre dos puntos separados por 100 (m)en 10 (s). Si la rapidez al pasar por el segundo punto es de 30 (m/s).¿cual es su aceleración?

  a = -1,33 m / s²

Explanation:

This problem is kinematic in one dimension, let's use the equations

         v = v₀ + a t

         v₀ = v - at

          x = v₀ t + ½ a t²

we substitute the first in the second equation

         x = (v-at) t + ½ a t²

         x -v t = a t² (1 + ½)

         a = (x -vt) / t² (1.5)

we calculate

         a = (100 - 30 10) / 10² (1.5) =-200 / 150

         a = -1,33 m / s²

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000