A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by (a) What are and , the

Question

A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by (a) What are and , the x- and y-components of the velocity of the car as functions of time? (b) What are the magnitude and direction of the velocity of the car at ? (b) What are the magnitude and direction of the acceleration of the car at ? calculus based

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Thu Thảo 5 years 2021-08-27T22:22:42+00:00 1 Answers 128 views 0

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  1. bonexptip
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    2021-08-27T22:23:57+00:00
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    Answer:

    a) ax = (-0.036a)t m/s^3; ay = 0.55 m/s^2

    b) |v| = 7.469 m/s; Θ = 59°

    c) |a| = 0.62 m/s^2; Θ = 298°

    Explanation:

    a)

    the instantaneous acceleration equation will be equal to:

    a = dv/dt

    ax(t) = (d*(5-0.018*t^2))/dt = -0.036*t

    ay(t) = (d*(2+0.55*t))/dt = 0.55

    a = (-0.036*t)i + 0.55j

    b)

    at a time of 8 seconds, the speed of the vector will be equal to:

    v = (5 – 0.018*8^2)i + (2 + 0.55*8)j = 3.85i + 6.4j

    the magnitude of the vector will be equal to:

    |v| = (vx^2 + vy^2)^1/2 = (3.85^2 + 6.4^2)^1/2 = 7.469 m/s

    the direction of the vector is equal to:

    Θ = tan-1(vy/vx) = tan-1(6.4/3.85) = 59°

    c)

    the vector acceleration at a time of 8 seconds will be equal to:

    a = (-0.036*8)i + 0.55j = -0.288i + 0.55j

    the magnitude will be equal to:

    |a| = (ax^2 + ay^2)^1/2 = ((-0.288^2) + 0.55^2)^1/2 = 0.62 m/s^2

    the direction:

    Θ = tan-1(ay/ax) = tan-1(0.55/-0.288) = 298°

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