A red toy car has an acceleration of 4.00 m/s², whereas a blue toy car has an acceleration of 2.00 m/s². You start the blue car from rest do

Question

A red toy car has an acceleration of 4.00 m/s², whereas a blue toy car has an acceleration of 2.00 m/s². You start the blue car from rest down a track of length 9.00 m. How much time must you wait before starting the red car from rest, such that the two cars reach the end of the track at the same time?

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Thu Cúc 5 years 2021-07-21T18:42:55+00:00 1 Answers 53 views 0

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  1. Thunguyet
    0
    2021-07-21T18:44:26+00:00
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    Answer:

    One must wait 0.88 secs before starting the red car

    Explanation:

    To calculate how much time to wait before starting the red car from rest, we will calculate the time it will take the blue car to reach the end of the track from rest and the time it will take the red car to reach the end of the track from rest; then, the difference between these times is the time one must wait before starting the red car from rest such that two cars reach the end of the track at the same time

    For the red car,

    Acceleration = 4.00 m/s²

    Initial velocity = 0 m/s (Since it is starting from rest)

    Distance = 9.00 m

    Let the time spent by the red car be t_{r}

    From one of the equations of kinematics for linear motion

    S = ut + \frac{1}{2}at^{2}

    Where

    S is the distance traveled

    u is the initial velocity

    t is the time

    and a is the acceleration

    Then, for the red car

    9.00 = (0)(t_{r}) + \frac{1}{2}(4.00)(t_{r})^{2}

    9.00 = (2.00)(t_{r})^{2}

    t_{r} = \sqrt{\frac{9.00}{2.00} }

    t_{r} =2.12 secs

    This is the time it will take the red car to reach the end of the track

    For the blue car

    Acceleration = 2.00 m/s²

    Initial velocity = 0 m/s (Since it is starting from rest)

    Distance = 9.00 m

    Let the time spent by the red car be t_{b}

    Also from

    S = ut + \frac{1}{2}at^{2}

    9.00 = (0)(t_{r}) + \frac{1}{2}(2.00)(t_{b})^{2}

    9.00 = (t_{b})^{2}

    t_{b} = \sqrt{9}

    t_{b} = 3secs

    This is the time it will take the blue car to reach the end of the track

    The difference of the times is

    t_{b} - t_{r} = 3 secs – 2.12 secs

    = 0.88 secs

    Hence, one must wait 0.88 secs before starting the red car

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