A pull back toy car accelerates for 1.8 seconds and covers 4 m while speeding up after being released from rest.

Question

A pull back toy car accelerates for 1.8 seconds and covers 4 m while speeding up after being released from rest.

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MichaelMet 4 months 2021-09-05T12:14:42+00:00 1 Answers 0 views 0

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    2021-09-05T12:16:40+00:00

    Answer:

    Acceleration, a = 2.47 seconds

    Explanation:

    Given the following data;

    Time, t = 1.8 seconds

    Distance = 4 meters

    Initial velocity = 0 m/s (since it’s starting from rest)

    To find the acceleration of the toy car, we would use the second equation of motion;

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

    Where;

    S represents the displacement or height measured in meters.

    u represents the initial velocity measured in meters per seconds.

    t represents the time measured in seconds.

    a represents acceleration measured in meters per seconds square.

    Substituting into the equation, we have;

    Substituting into the formula, we have;

     4 = 0(1.8) + \frac {1}{2}*a*(1.8)^{2}

     4 = 0 + \frac {1}{2}*a*3.24

     4 = \frac {1}{2}*a*3.24

    Cross multiplying, we have;

     8 = 3.24a

     a = \frac {8}{3.24}

    Acceleration, a = 2.47 seconds

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