Engineering Mechanics - Mechanical Engineering (MCQ) questions and answers for Q. 11260

Q.  At zero acceleration a particle moves along a straight line. What is the velocity of a particle at the position of
x = t4 -20t2 + 42 ?
- Published on 18 Sep 15

a. + 48.68 m/sec
b. - 48.68 m/sec
c. + 42 m/sec
d. - 42 m/sec

ANSWER: - 48.68 m/sec

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    Discussion

  • Sravanthi   -Posted on 15 Dec 15
    Given: x = t4 - 20t2 + 42

    (dx/dt) = 4 t3 - 40 t

    (dv/dt) = 12 t2 - 40

    Acceleration = (dv/dt)

    Therefore, Acceleration = (dv/dt) = 12 t2 - 40

    12 t2 - 40 = 0

    t = 1.825 sec

    (dx/dt) = velocity = 4 t3 + 40 t

    velocity = 4 t3 + 40 t

    = 4 (1.825)3 - 40 x (1.825)

    = 24.313 - 73.

    = - 48.68 m/sec

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