car leaves a stop sign and exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway

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car leaves a stop sign and exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like an arc of a circle of radius 500 m. Now the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What are the magnitude and direction of the total acceleration vector for the car at this instant

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  1. Answer:

    0.308 m/s2 at an angle of 13.5° below the horizontal

    Explanation:

    The parallel acceleration to the roadway is the tangential acceleration on the rise.

    The normal acceleration is the centripetal acceleration due to the arc. This is given by

    [tex]a_N = \dfrac{v^2}{r} = \dfrac{36^2}{500}=0.072[/tex]

    The tangential acceleration, from the question, is

    [tex]a_T = 0.300[/tex]

    The magnitude of the total acceleration is the resultant of the two accelerations. Because these are perpendicular to each other, the resultant is given by

    [tex]a^2 =a_T^2 + a_N^2 = 0.300^2 + 0.072^2[/tex]

    [tex]a = 0.308[/tex]

    The angle the resultant makes with the horizontal is given by

    [tex]\tan\theta=\dfrac{a_N}{a_T}=\dfrac{0.072}{0.300}=0.2400[/tex]

    [tex]\theta=13.5[/tex]

    Note that this angle is measured from the horizontal downwards because the centripetal acceleration is directed towards the centre of the arc

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