En una competencia de autos robots, la prueba consiste en pasar por 4 puntos distintos ubicados sobre una línea
recta, donde se harán mediciones de tiempo. Uno de los
participantes recolecta información para saber si su
robot es viable para la competencia. Realiza un ensayo
en una pista similar a la de la competencia. En el punto
1, el robot parte del reposo y cambia su velocidad a
razón de 10 cm/s². Desde el punto 2 al 3 hay 105 cm y
se demora 3 s en cubrir dicha distancia. Termina su
recorrido en el punto 4, teniendo presente que entre el
punto 3 y el 4 hay 55 cm. El dueño del robot desea
saber:
a. La velocidad del robot en los puntos 2,3 y 4
b. La distancia desde al punto 1 hasta el 2
c. Cuanto se demoró en el espacio de 1 a 2 y en el de 3
a 4

a) The speeds of the robot at points 2, 3 and 4 are 20 cm/s, 50 cm/s and 60 cm/s

b) The distance from point 1 to point 2 = 20 cm

c) How long did it take in the space of 1 to 2 and in the space of 3 to 4.

t₁₂ = 2 s

t₃₄ = 1 s

Explanation:

We will use the equations of motion to solve this question

At point 1, the robot starts from rest and changes its speed at a rate of 10 cm / s²

Let the speed at points 1, 2, 3 and 4 be v₁, v₂, v₃ and v₄

Since the robot starts at rest from the point I, v₁ = 0 cm/s

The robot is said to accelerate uniformly at 10 cm/s²

v₂ = v₁ + at₁₂

v₁ = 0 cm/s (the robot starts from rest at point 1)

a = 10 cm/s

t₁₂ = time taken to cover the distance between 1 and 2

v₂ = 0 + 10×t₁₂ = (10t₁₂) cm/s

Point 2 to 3 was 105 cm and this took the robot 3 s to complete

v₂ = ?

v₃ = ?

x₂₃ = distance between points 2 and 3 = 105 cm

t₂₃ = time taken to cover the distance between point 2 and 3 = 3 s

a = acceleration of the car, still constant = 10 cm/s²

x₂₃ = v₂t₂₃ + ½at₂₃²

105 = (v₂×3) + 0.5×10×3²

105 = 3v₂ + 45

3v₂ = 105 - 45 = 60

v₂ = (60/3) = 20 cm/s = 0.2 m/s

Applying equations of motion again

v₃ = v₂ + at₂₃

v₃ = 20 + 10×3 = 20 + 30 = 50 cm/s = 0.5 m/s

For points 3 to 4

v₃ = 50 cm/s

x₃₄ = 55 cm

a = 10 cm/s²

v₄ = ?

Another equation of motion

(v₄)² = (v₃)² + 2ax₃₄

(v₄)² = 50² + 2×10×55 = 2500 + 1100 = 3600

v₄ = √3600 = 60 cm/s = 0.6 m/s

b) To calculate the distance from 1 to 2, we write all we know about 1 to 2

v₁ = 0 cm/s

v₂ = 20 cm/s

a = 10 cm/s²

x₁₂ = ?

(v₂)² = (v₁)² + 2ax₁₂

20² = 0² + (2×10×x₁₂)

20x₁₂ = 400

x₁₂ = (400/20) = 20 cm

c) To find the time taken to complete points 1 to 2

v₁ = 0 cm/s

v₂ = 20 cm/s

a = 10 cm/s²

t₁₂ = ?

v₂= v₁ + at₁₂

20 = 0 + 10t₁₂

t₁₂ = (20/10) = 2 s

For the points 3 to 4

v₃ = 50 cm/s

v₄ = 60 cm/s

a = 10 cm/s²

t₃₄ = ?

v₄ = v₃ + at₃₄

60 = 50 + 10t₃₄

10t₃₄ = 10

t₃₄ = (10/10) = 1 s

Hope this Helps!!!

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