En una competencia de autos robots, la prueba consiste en pasar por 4 puntos distintos ubicados sobre una línea recta, donde se

Question

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

in progress 0
Doris 5 years 2021-09-01T01:22:18+00:00 1 Answers 17 views 0

Answers ( )

  1. Nem
    0
    2021-09-01T01:23:33+00:00
    Reply

    Answer:

    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!!!

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