A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 25.0 m

Question

A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 25.0 m/s. Then the vehicle travels for 39.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.
(a) How long is the self-driving car in motion (in s)?
(b) What is the average velocity of the self-driving car for the motion described? (Enter the magnitude in m/s.) m/s

in progress 0
Gerda 3 years 2021-08-02T01:38:41+00:00 1 Answers 39 views 0

Answers ( )

  1. minhkhue
    0
    2021-08-02T01:40:10+00:00
    Reply

    Answer:

    56.5\ \text{s}

    21.13\ \text{m/s}

    Explanation:

    v = Final velocity

    u = Initial velocity

    a = Acceleration

    t = Time

    s = Displacement

    Here the kinematic equations of motion are used

    v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}

    Time the car is at constant velocity is 39 s

    Time the car is decelerating is 5 s

    Total time the car is in motion is 12.5+39+5=56.5\ \text{s}

    Distance traveled

    v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{25^2-0}{2\times 2}\\\Rightarrow s=156.25\ \text{m}

    s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}

    v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{5}\\\Rightarrow a=-5\ \text{m/s}^2

    s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-25^2}{2\times -5}\\\Rightarrow s=62.5\ \text{m}

    The total displacement of the car is 156.25+975+62.5=1193.75\ \text{m}

    Average velocity is given by

    \dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}

    The average velocity of the car is 21.13\ \text{m/s}.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )